Dari Wikibuku bahasa Indonesia, sumber buku teks bebas
contoh soal
Berapa hasil dari
2015
⋅
2017
⋅
2023
⋅
2025
+
64
{\displaystyle {\sqrt {2015\cdot 2017\cdot 2023\cdot 2025+64}}}
Jawaban
Misalkan 2020 = p
2015
⋅
2017
⋅
2023
⋅
2025
+
64
=
(
2020
−
5
)
⋅
(
2020
−
3
)
⋅
(
2020
+
3
)
⋅
(
2020
+
5
)
+
64
=
(
p
−
5
)
⋅
(
p
−
3
)
⋅
(
p
+
3
)
⋅
(
p
+
5
)
+
64
=
(
p
−
5
)
⋅
(
p
+
5
)
⋅
(
p
−
3
)
⋅
(
p
+
3
)
+
64
=
(
p
2
−
25
)
⋅
(
p
2
−
9
)
+
64
=
p
4
−
34
p
2
+
225
+
64
=
p
4
−
34
p
2
+
289
=
(
p
2
−
17
)
2
=
p
2
−
17
=
2020
2
−
17
=
(
2000
+
20
)
2
−
17
=
4.000.000
+
80.000
+
400
−
17
=
4.080.383
{\displaystyle {\begin{aligned}{\text{Misalkan 2020 = p}}\\{\sqrt {2015\cdot 2017\cdot 2023\cdot 2025+64}}&={\sqrt {(2020-5)\cdot (2020-3)\cdot (2020+3)\cdot (2020+5)+64}}\\&={\sqrt {(p-5)\cdot (p-3)\cdot (p+3)\cdot (p+5)+64}}\\&={\sqrt {(p-5)\cdot (p+5)\cdot (p-3)\cdot (p+3)+64}}\\&={\sqrt {(p^{2}-25)\cdot (p^{2}-9)+64}}\\&={\sqrt {p^{4}-34p^{2}+225+64}}\\&={\sqrt {p^{4}-34p^{2}+289}}\\&={\sqrt {(p^{2}-17)^{2}}}\\&=p^{2}-17\\&=2020^{2}-17\\&=(2000+20)^{2}-17\\&=4.000.000+80.000+400-17\\&=4.080.383\\\end{aligned}}}
Berapa nilai x dari
x
2
−
x
−
x
2
−
x
−
x
2
−
x
−
…
x
2
x
2
x
2
…
3
3
3
=
9
10
{\displaystyle {\frac {\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {\dots }}}}}}}}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}\dots }}}}}}}={\frac {9}{10}}}
Jawaban
x
2
−
x
−
x
2
−
x
−
x
2
−
x
−
…
x
2
x
2
x
2
…
3
3
3
=
9
10
misalkan untuk
x
2
−
x
−
x
2
−
x
−
x
2
−
x
−
…
=
p
x
2
−
x
−
x
2
−
x
−
x
2
−
x
−
…
=
p
x
2
−
x
−
x
2
−
x
−
x
2
−
x
−
…
=
p
2
x
2
−
x
−
p
=
p
2
x
2
−
2
x
+
1
+
x
−
1
=
p
2
+
p
(
x
−
1
)
2
+
(
x
−
1
)
=
p
2
+
p
x
−
1
=
p
misalkan untuk
x
2
x
2
x
2
…
3
3
3
=
q
x
2
x
2
x
2
…
3
3
3
=
q
x
2
x
2
x
2
…
3
3
=
q
3
x
2
q
=
q
3
x
2
=
q
2
x
=
q
x
−
1
x
=
9
10
x
=
10
{\displaystyle {\begin{aligned}{\frac {\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {\dots }}}}}}}}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}\dots }}}}}}}&={\frac {9}{10}}\\{\text{misalkan untuk }}{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {\dots }}}}}}}}=p\\{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {\dots }}}}}}}}&=p\\x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {\dots }}}}}}&=p^{2}\\x^{2}-x-p&=p^{2}\\x^{2}-2x+1+x-1&=p^{2}+p\\(x-1)^{2}+(x-1)&=p^{2}+p\\x-1&=p\\{\text{misalkan untuk }}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}\dots }}}}}}&=q\\{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}\dots }}}}}}&=q\\x^{2}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}\dots }}}}&=q^{3}\\x^{2}q&=q^{3}\\x^{2}&=q^{2}\\x&=q\\{\frac {x-1}{x}}&={\frac {9}{10}}\\x&=10\\\end{aligned}}}
Berapa nilai x dari
(
x
x
+
10
)
x
+
10
=
1
1024
{\displaystyle ({\frac {x}{x+10}})^{x+10}={\frac {1}{1024}}}
Jawaban
(
x
+
10
x
)
−
(
x
+
10
)
=
(
1024
)
−
1
(
x
+
10
x
)
x
+
10
=
1024
(
x
+
10
x
)
x
+
10
=
2
10
(
x
+
10
x
)
x
+
10
10
=
2
(
1
+
10
x
)
1
+
x
10
=
2
(
1
+
10
x
)
1
+
x
10
=
(
1
2
)
−
1
(
1
+
10
x
)
1
+
x
10
=
(
1
+
(
−
1
2
)
)
(
1
+
(
−
2
1
)
)
10
x
=
−
1
2
x
=
−
20
{\displaystyle {\begin{aligned}({\frac {x+10}{x}})^{-(x+10)}&=(1024)^{-1}\\({\frac {x+10}{x}})^{x+10}&=1024\\({\frac {x+10}{x}})^{x+10}&=2^{10}\\({\frac {x+10}{x}})^{\frac {x+10}{10}}&=2\\(1+{\frac {10}{x}})^{1+{\frac {x}{10}}}&=2\\(1+{\frac {10}{x}})^{1+{\frac {x}{10}}}&=({\frac {1}{2}})^{-1}\\(1+{\frac {10}{x}})^{1+{\frac {x}{10}}}&=(1+(-{\frac {1}{2}}))^{(1+(-{\frac {2}{1}}))}\\{\frac {10}{x}}&=-{\frac {1}{2}}\\x&=-20\\\end{aligned}}}
Berapa nilai x dari
x
+
x
+
1
2
+
x
+
1
4
=
4
{\displaystyle x+{\sqrt {x+{\frac {1}{2}}+{\sqrt {x+{\frac {1}{4}}}}}}=4}
Jawaban
x
+
1
2
+
x
+
1
4
=
(
x
+
1
4
)
2
+
2
⋅
x
+
1
4
⋅
1
2
+
(
1
2
)
2
=
(
x
+
1
4
+
1
2
)
2
x
+
x
+
1
2
+
x
+
1
4
=
4
x
+
(
x
+
1
4
+
1
2
)
2
=
4
x
+
x
+
1
4
+
1
2
=
4
(
x
+
1
4
+
1
2
)
2
=
4
x
+
1
4
+
1
2
=
2
x
+
1
4
=
3
2
x
+
1
4
=
9
4
x
=
2
{\displaystyle {\begin{aligned}x+{\frac {1}{2}}+{\sqrt {x+{\frac {1}{4}}}}&=({\sqrt {x+{\frac {1}{4}}}})^{2}+2\cdot {\sqrt {x+{\frac {1}{4}}}}\cdot {\frac {1}{2}}+({\frac {1}{2}})^{2}\\&=({\sqrt {x+{\frac {1}{4}}}}+{\frac {1}{2}})^{2}\\x+{\sqrt {x+{\frac {1}{2}}+{\sqrt {x+{\frac {1}{4}}}}}}&=4\\x+{\sqrt {({\sqrt {x+{\frac {1}{4}}}}+{\frac {1}{2}})^{2}}}&=4\\x+{\sqrt {x+{\frac {1}{4}}}}+{\frac {1}{2}}&=4\\({\sqrt {x+{\frac {1}{4}}}}+{\frac {1}{2}})^{2}&=4\\{\sqrt {x+{\frac {1}{4}}}}+{\frac {1}{2}}&=2\\{\sqrt {x+{\frac {1}{4}}}}&={\frac {3}{2}}\\x+{\frac {1}{4}}&={\frac {9}{4}}\\x&=2\\\end{aligned}}}
Berapa nilai x dari
x
3
8
−
x
2
+
x
2
−
8
=
0
{\displaystyle {\frac {x^{3}}{\sqrt {8-x^{2}}}}+x^{2}-8=0}
Jawaban
x
3
8
−
x
2
+
x
2
−
8
=
0
x
3
8
−
x
2
=
8
−
x
2
x
3
=
(
8
−
x
2
)
3
2
x
=
(
8
−
x
2
)
1
2
x
2
=
8
−
x
2
2
x
2
−
8
=
0
x
2
−
4
=
0
(
x
−
2
)
(
x
+
2
)
=
0
membuktikan
x
=
2
maka hasilnya 0
x
=
−
2
maka hasilnya -8
jadi
x
=
2
{\displaystyle {\begin{aligned}{\frac {x^{3}}{\sqrt {8-x^{2}}}}+x^{2}-8&=0\\{\frac {x^{3}}{\sqrt {8-x^{2}}}}&=8-x^{2}\\x^{3}&=(8-x^{2})^{\frac {3}{2}}\\x&=(8-x^{2})^{\frac {1}{2}}\\x^{2}&=8-x^{2}\\2x^{2}-8&=0\\x^{2}-4&=0\\(x-2)(x+2)&=0\\{\text{membuktikan }}\\x=2{\text{ maka hasilnya 0 }}\\x=-2{\text{ maka hasilnya -8 }}\\{\text{jadi }}x=2\\\end{aligned}}}
Berapa nilai x dari
3
x
+
5
+
4
x
+
5
=
x
{\displaystyle {\sqrt {3x+5+{\sqrt {4x+5}}}}=x}
Jawaban
3
x
+
5
+
4
x
+
5
=
x
4
x
+
5
+
4
x
+
5
−
x
=
x
misalkan
4
x
+
5
=
y
dan
4
x
+
5
=
y
2
4
x
+
5
+
4
x
+
5
−
x
=
x
y
2
+
y
−
x
=
x
y
2
+
y
=
x
2
+
x
y
=
x
4
x
+
5
=
y
2
4
x
+
5
=
x
2
x
2
−
4
x
−
5
=
0
(
x
−
5
)
(
x
+
1
)
=
0
x
=
5
atau
x
=
−
1
(TM)
{\displaystyle {\begin{aligned}{\sqrt {3x+5+{\sqrt {4x+5}}}}&=x\\{\sqrt {4x+5+{\sqrt {4x+5}}-x}}&=x\\{\text{misalkan }}{\sqrt {4x+5}}=y{\text{ dan }}4x+5=y^{2}\\{\sqrt {4x+5+{\sqrt {4x+5}}-x}}&=x\\{\sqrt {y^{2}+y-x}}&=x\\y^{2}+y&=x^{2}+x\\y=x\\4x+5&=y^{2}\\4x+5&=x^{2}\\x^{2}-4x-5&=0\\(x-5)(x+1)&=0\\x=5&{\text{ atau }}x=-1{\text{ (TM) }}\\\end{aligned}}}
Berapa nilai x dari
1
+
1
+
x
=
x
3
{\displaystyle {\sqrt {1+{\sqrt {1+x}}}}={\sqrt[{3}]{x}}}
Jawaban
1
+
1
+
x
=
x
3
x
3
=
n
x
=
n
3
1
+
1
+
n
3
=
n
1
+
1
+
n
3
=
n
2
1
+
n
3
=
n
2
−
1
1
+
n
3
=
n
4
−
2
n
2
+
1
n
4
−
n
3
−
2
n
2
=
0
n
2
(
n
2
−
n
−
2
)
=
0
n
2
(
n
−
2
)
(
n
+
1
)
=
0
n
=
0
,
n
=
2
atau
n
=
−
1
n
=
0
x
=
0
3
=
0
n
=
2
x
=
2
3
=
8
n
=
−
1
x
=
(
−
1
)
3
=
−
1
yang paling mungkin untuk nilai x adalah
8
{\displaystyle {\begin{aligned}{\sqrt {1+{\sqrt {1+x}}}}&={\sqrt[{3}]{x}}\\{\sqrt[{3}]{x}}&=n\\x&=n^{3}\\{\sqrt {1+{\sqrt {1+n^{3}}}}}&=n\\1+{\sqrt {1+n^{3}}}&=n^{2}\\{\sqrt {1+n^{3}}}&=n^{2}-1\\1+n^{3}&=n^{4}-2n^{2}+1\\n^{4}-n^{3}-2n^{2}&=0\\n^{2}(n^{2}-n-2)&=0\\n^{2}(n-2)(n+1)&=0\\n=0,n=2{\text{ atau }}n=-1\\n&=0\\x&=0^{3}\\&=0\\n&=2\\x&=2^{3}\\&=8\\n&=-1\\x&=(-1)^{3}\\&=-1\\{\text{yang paling mungkin untuk nilai x adalah }}8\\\end{aligned}}}
Berapa nilai x dari
1
+
x
+
x
1
+
x
−
x
=
1
+
x
x
{\displaystyle {\frac {{\sqrt {1+x}}+{\sqrt {x}}}{{\sqrt {1+x}}-{\sqrt {x}}}}={\frac {\sqrt {1+x}}{\sqrt {x}}}}
Jawaban
1
+
x
+
x
1
+
x
−
x
=
1
+
x
x
x
(
1
+
x
+
x
)
=
(
1
+
x
−
x
)
1
+
x
x
(
1
+
x
)
+
x
=
1
+
x
−
x
(
1
+
x
)
2
x
(
1
+
x
)
=
1
x
(
1
+
x
)
=
1
2
x
(
1
+
x
)
=
1
4
x
2
+
x
=
1
4
4
x
2
+
4
x
=
1
4
x
2
+
4
x
−
1
=
0
x
=
−
4
±
4
2
−
4
(
4
)
(
−
1
)
2
(
4
)
=
−
4
±
32
8
=
−
4
±
4
2
8
=
−
1
±
2
2
karena akar x harus minimal nol jadi
x
=
−
1
+
2
2
{\displaystyle {\begin{aligned}{\frac {{\sqrt {1+x}}+{\sqrt {x}}}{{\sqrt {1+x}}-{\sqrt {x}}}}&={\frac {\sqrt {1+x}}{\sqrt {x}}}\\{\sqrt {x}}({\sqrt {1+x}}+{\sqrt {x}})&=({\sqrt {1+x}}-{\sqrt {x}}){\sqrt {1+x}}\\{\sqrt {x(1+x)}}+x&=1+x-{\sqrt {x(1+x)}}\\2{\sqrt {x(1+x)}}&=1\\{\sqrt {x(1+x)}}&={\frac {1}{2}}\\x(1+x)&={\frac {1}{4}}\\x^{2}+x&={\frac {1}{4}}\\4x^{2}+4x&=1\\4x^{2}+4x-1&=0\\x&={\frac {-4\pm {\sqrt {4^{2}-4(4)(-1)}}}{2(4)}}\\&={\frac {-4\pm {\sqrt {32}}}{8}}\\&={\frac {-4\pm 4{\sqrt {2}}}{8}}\\&={\frac {-1\pm {\sqrt {2}}}{2}}\\{\text{karena akar x harus minimal nol jadi }}x={\frac {-1+{\sqrt {2}}}{2}}\\\end{aligned}}}
Berapa nilai x dari
x
+
x
2
−
1
x
−
x
2
−
1
+
x
−
x
2
−
1
x
+
x
2
−
1
=
98
{\displaystyle {\frac {x+{\sqrt {x^{2}-1}}}{x-{\sqrt {x^{2}-1}}}}+{\frac {x-{\sqrt {x^{2}-1}}}{x+{\sqrt {x^{2}-1}}}}=98}
Jawaban
x
+
x
2
−
1
x
−
x
2
−
1
+
x
−
x
2
−
1
x
+
x
2
−
1
=
98
misalkan
x
2
−
1
=
n
x
+
n
x
−
n
+
x
−
n
x
+
n
=
98
(
x
+
n
)
2
+
(
x
−
n
)
2
(
x
−
n
)
(
x
+
n
)
=
98
x
2
+
2
x
n
+
n
2
+
x
2
−
2
x
n
+
n
2
x
2
−
n
2
=
98
2
(
x
2
+
n
2
)
x
2
−
n
2
=
98
x
2
+
n
2
x
2
−
n
2
=
49
x
2
+
n
2
=
49
(
x
2
−
n
2
)
x
2
+
n
2
=
49
x
2
−
49
n
2
48
x
2
=
50
n
2
24
x
2
=
25
n
2
24
x
2
=
25
(
x
2
−
1
)
2
24
x
2
=
25
(
x
2
−
1
)
24
x
2
=
25
x
2
−
25
x
2
=
25
x
=
±
5
{\displaystyle {\begin{aligned}{\frac {x+{\sqrt {x^{2}-1}}}{x-{\sqrt {x^{2}-1}}}}+{\frac {x-{\sqrt {x^{2}-1}}}{x+{\sqrt {x^{2}-1}}}}&=98\\{\text{misalkan }}{\sqrt {x^{2}-1}}=n\\{\frac {x+n}{x-n}}+{\frac {x-n}{x+n}}&=98\\{\frac {(x+n)^{2}+(x-n)^{2}}{(x-n)(x+n)}}&=98\\{\frac {x^{2}+2xn+n^{2}+x^{2}-2xn+n^{2}}{x^{2}-n^{2}}}&=98\\{\frac {2(x^{2}+n^{2})}{x^{2}-n^{2}}}&=98\\{\frac {x^{2}+n^{2}}{x^{2}-n^{2}}}&=49\\x^{2}+n^{2}&=49(x^{2}-n^{2})\\x^{2}+n^{2}&=49x^{2}-49n^{2}\\48x^{2}&=50n^{2}\\24x^{2}&=25n^{2}\\24x^{2}&=25({\sqrt {x^{2}-1}})^{2}\\24x^{2}&=25(x^{2}-1)\\24x^{2}&=25x^{2}-25\\x^{2}&=25\\x&=\pm 5\\\end{aligned}}}
Berapa nilai x dari
x
+
x
−
x
−
x
=
5
4
x
x
+
x
{\displaystyle {\sqrt {x+{\sqrt {x}}}}-{\sqrt {x-{\sqrt {x}}}}={\frac {5}{4}}{\sqrt {\frac {x}{x+{\sqrt {x}}}}}}
Jawaban
misalkan
x
=
y
dan
x
=
y
2
x
+
x
−
x
−
x
=
5
4
x
x
+
x
y
2
+
y
−
y
2
−
y
=
5
4
y
2
y
2
+
y
y
2
+
y
−
y
2
−
y
=
5
4
y
y
2
+
y
y
2
+
y
−
(
y
2
+
y
)
(
y
2
−
y
)
=
5
4
y
y
2
+
y
−
y
4
−
y
2
=
5
4
y
y
2
+
y
−
y
2
(
y
2
−
1
)
=
5
4
y
y
(
y
+
1
)
−
y
y
2
−
1
=
5
4
y
y
+
1
−
y
2
−
1
=
5
4
−
y
2
−
1
=
1
4
−
y
y
2
−
1
=
(
1
4
−
y
)
2
y
2
−
1
=
1
16
−
1
2
y
+
y
2
−
1
=
1
16
−
1
2
y
1
2
y
=
1
16
+
1
1
2
y
=
17
16
y
=
17
8
x
=
(
17
8
)
2
=
289
64
{\displaystyle {\begin{aligned}{\text{misalkan }}{\sqrt {x}}=y{\text{ dan }}x=y^{2}\\{\sqrt {x+{\sqrt {x}}}}-{\sqrt {x-{\sqrt {x}}}}&={\frac {5}{4}}{\sqrt {\frac {x}{x+{\sqrt {x}}}}}\\{\sqrt {y^{2}+y}}-{\sqrt {y^{2}-y}}&={\frac {5}{4}}{\sqrt {\frac {y^{2}}{y^{2}+y}}}\\{\sqrt {y^{2}+y}}-{\sqrt {y^{2}-y}}&={\frac {5}{4}}{\frac {y}{\sqrt {y^{2}+y}}}\\y^{2}+y-{\sqrt {(y^{2}+y)(y^{2}-y)}}&={\frac {5}{4}}y\\y^{2}+y-{\sqrt {y^{4}-y^{2}}}&={\frac {5}{4}}y\\y^{2}+y-{\sqrt {y^{2}(y^{2}-1)}}&={\frac {5}{4}}y\\y(y+1)-y{\sqrt {y^{2}-1}}&={\frac {5}{4}}y\\y+1-{\sqrt {y^{2}-1}}&={\frac {5}{4}}\\-{\sqrt {y^{2}-1}}&={\frac {1}{4}}-y\\y^{2}-1&=({\frac {1}{4}}-y)^{2}\\y^{2}-1&={\frac {1}{16}}-{\frac {1}{2}}y+y^{2}\\-1&={\frac {1}{16}}-{\frac {1}{2}}y\\{\frac {1}{2}}y&={\frac {1}{16}}+1\\{\frac {1}{2}}y&={\frac {17}{16}}\\y&={\frac {17}{8}}\\x&=({\frac {17}{8}})^{2}\\&={\frac {289}{64}}\\\end{aligned}}}
Berapa nilai x dari
(
8
+
x
)
2
3
−
(
8
+
x
)
(
27
−
x
)
3
+
(
27
−
x
)
2
3
=
7
{\displaystyle {\sqrt[{3}]{(8+x)^{2}}}-{\sqrt[{3}]{(8+x)(27-x)}}+{\sqrt[{3}]{(27-x)^{2}}}=7}
Jawaban
(
8
+
x
)
2
3
−
(
8
+
x
)
(
27
−
x
)
3
+
(
27
−
x
)
2
3
=
7
(
8
+
x
3
)
2
−
8
+
x
3
27
−
x
3
+
(
27
−
x
3
)
2
=
7
misalkan
8
+
x
3
=
a
,
8
+
x
=
a
3
,
27
−
x
3
=
b
dan
27
−
x
=
b
3
a
2
−
a
b
+
b
2
=
7
a
3
+
b
3
=
8
+
x
+
27
−
x
=
35
a
3
+
b
3
=
(
a
+
b
)
(
a
2
−
a
b
+
b
2
)
35
=
(
a
+
b
)
(
7
)
a
+
b
=
5
b
=
5
−
a
(
a
+
b
)
3
=
a
3
+
b
3
+
3
a
b
(
a
+
b
)
5
3
=
35
+
3
a
b
(
5
)
125
=
35
+
15
a
b
80
=
15
a
b
a
b
=
6
a
(
5
−
a
)
=
6
5
a
−
a
2
=
6
a
2
−
5
a
+
6
=
6
(
a
−
2
)
(
a
−
3
)
=
6
a
=
2
atau
a
=
3
a
=
2
,
b
=
3
dan
a
=
3
,
b
=
2
8
+
x
=
a
3
=
2
3
=
8
x
=
0
8
+
x
=
a
3
=
3
3
=
27
x
=
19
{\displaystyle {\begin{aligned}{\sqrt[{3}]{(8+x)^{2}}}-{\sqrt[{3}]{(8+x)(27-x)}}+{\sqrt[{3}]{(27-x)^{2}}}&=7\\({\sqrt[{3}]{8+x}})^{2}-{\sqrt[{3}]{8+x}}{\sqrt[{3}]{27-x}}+({\sqrt[{3}]{27-x}})^{2}&=7\\{\text{misalkan }}{\sqrt[{3}]{8+x}}=a,8+x=a^{3},{\sqrt[{3}]{27-x}}=b{\text{ dan }}27-x=b^{3}\\a^{2}-ab+b^{2}&=7\\a^{3}+b^{3}&=8+x+27-x\\&=35\\a^{3}+b^{3}&=(a+b)(a^{2}-ab+b^{2})\\35&=(a+b)(7)\\a+b&=5\\b&=5-a\\(a+b)^{3}&=a^{3}+b^{3}+3ab(a+b)\\5^{3}&=35+3ab(5)\\125&=35+15ab\\80&=15ab\\ab&=6\\a(5-a)&=6\\5a-a^{2}&=6\\a^{2}-5a+6&=6\\(a-2)(a-3)&=6\\a=2&{\text{ atau }}a=3\\a=2,b=3{\text{ dan }}a=3,b=2\\8+x&=a^{3}\\&=2^{3}\\&=8\\x&=0\\8+x&=a^{3}\\&=3^{3}\\&=27\\x&=19\\\end{aligned}}}
Berapa nilai x dari
3
x
+
5
x
−
9
x
+
15
x
−
25
x
=
1
{\displaystyle 3^{x}+5^{x}-9^{x}+15^{x}-25^{x}=1}
Jawaban
3
x
+
5
x
−
9
x
+
15
x
−
25
x
=
1
3
x
+
5
x
−
(
3
2
)
x
+
(
3
⋅
5
)
x
−
(
5
2
)
x
=
1
3
x
+
5
x
−
(
3
x
)
2
+
(
3
x
⋅
5
x
)
−
(
5
x
)
2
=
1
misalkan
3
x
=
a
dan
5
x
=
b
a
+
b
−
a
2
+
a
b
−
b
2
=
1
a
2
−
a
b
+
b
2
−
a
−
b
+
1
=
0
2
a
2
−
2
a
b
+
2
b
2
−
2
a
−
2
b
+
2
=
0
a
2
−
2
a
b
+
b
2
+
a
2
−
2
a
+
1
+
b
2
−
2
b
+
1
=
0
(
a
−
b
)
2
+
(
a
−
1
)
2
+
(
b
−
1
)
2
=
0
a
−
b
=
0
;
a
−
1
=
0
;
b
−
1
=
0
a
=
b
=
1
3
x
=
1
x
=
0
{\displaystyle {\begin{aligned}3^{x}+5^{x}-9^{x}+15^{x}-25^{x}&=1\\3^{x}+5^{x}-(3^{2})^{x}+(3\cdot 5)^{x}-(5^{2})^{x}&=1\\3^{x}+5^{x}-(3^{x})^{2}+(3^{x}\cdot 5^{x})-(5^{x})^{2}&=1\\{\text{misalkan }}3^{x}=a{\text{ dan }}5^{x}=b\\a+b-a^{2}+ab-b^{2}&=1\\a^{2}-ab+b^{2}-a-b+1&=0\\2a^{2}-2ab+2b^{2}-2a-2b+2&=0\\a^{2}-2ab+b^{2}+a^{2}-2a+1+b^{2}-2b+1&=0\\(a-b)^{2}+(a-1)^{2}+(b-1)^{2}&=0\\a-b=0;a-1=0;b-1&=0\\a=b&=1\\3^{x}&=1\\x&=0\\\end{aligned}}}
Berapa nilai x dari
6
l
o
g
x
2
+
6
x
l
o
g
6
x
=
1
{\displaystyle ^{6}logx^{2}+^{6x}log{\frac {6}{x}}=1}
Jawaban
6
l
o
g
x
2
+
6
x
l
o
g
6
x
=
1
misalkan
6
x
=
a
maka
x
=
a
6
6
l
o
g
x
2
+
6
x
l
o
g
6
x
=
1
6
l
o
g
(
a
6
)
2
+
6
a
6
l
o
g
6
a
6
=
1
6
l
o
g
a
2
6
2
+
a
l
o
g
6
2
a
=
1
6
l
o
g
a
2
−
6
l
o
g
6
2
+
a
l
o
g
6
2
−
a
l
o
g
a
=
1
2
6
l
o
g
a
−
2
6
l
o
g
6
+
2
a
l
o
g
6
−
a
l
o
g
a
=
1
2
6
l
o
g
a
−
2
+
2
1
6
l
o
g
a
−
1
=
1
2
6
l
o
g
a
+
2
1
6
l
o
g
a
−
4
=
0
2
6
l
o
g
2
a
−
4
6
l
o
g
a
+
2
=
0
6
l
o
g
2
a
−
2
6
l
o
g
a
+
1
=
0
(
6
l
o
g
a
−
1
)
2
=
0
6
l
o
g
a
=
1
a
=
6
x
=
a
6
=
6
6
=
1
{\displaystyle {\begin{aligned}^{6}logx^{2}+^{6x}log{\frac {6}{x}}&=1\\{\text{misalkan }}6x=a{\text{ maka }}x={\frac {a}{6}}\\^{6}logx^{2}+^{6x}log{\frac {6}{x}}&=1\\^{6}log({\frac {a}{6}})^{2}+^{6{\frac {a}{6}}}log{\frac {6}{\frac {a}{6}}}&=1\\^{6}log{\frac {a^{2}}{6^{2}}}+^{a}log{\frac {6^{2}}{a}}&=1\\^{6}loga^{2}-^{6}log6^{2}+^{a}log6^{2}-^{a}loga&=1\\2^{6}loga-2^{6}log6+2^{a}log6-^{a}loga&=1\\2^{6}loga-2+2{\frac {1}{^{6}loga}}-1&=1\\2^{6}loga+2{\frac {1}{^{6}loga}}-4&=0\\2^{6}log^{2}a-4^{6}loga+2&=0\\^{6}log^{2}a-2^{6}loga+1&=0\\(^{6}loga-1)^{2}&=0\\^{6}loga&=1\\a&=6\\x&={\frac {a}{6}}\\&={\frac {6}{6}}\\&=1\\\end{aligned}}}
Berapa nilai x dari (x+500)3 +x=20? Jawaban
(
x
+
500
)
3
+
x
=
20
misalkan
a
=
x
+
500
maka
x
=
a
−
500
a
3
+
a
−
500
=
20
a
3
+
a
=
520
a
(
a
2
+
1
)
=
8
⋅
65
a
(
a
2
+
1
)
=
8
(
64
+
1
)
a
(
a
2
+
1
)
=
8
(
8
2
+
1
)
a
=
8
x
=
8
−
500
=
−
492
{\displaystyle {\begin{aligned}(x+500)^{3}+x&=20\\{\text{misalkan }}a=x+500{\text{ maka }}x=a-500\\a^{3}+a-500&=20\\a^{3}+a&=520\\a(a^{2}+1)&=8\cdot 65\\a(a^{2}+1)&=8(64+1)\\a(a^{2}+1)&=8(8^{2}+1)\\a&=8\\x&=8-500\\&=-492\\\end{aligned}}}
Berapa nilai x dari
x
n
+
4
n
x
n
+
16
n
n
−
1
2
=
0
{\displaystyle {\sqrt[{n}]{\frac {x^{n}+4^{n}}{x^{n}+16^{n}}}}-{\frac {1}{2}}=0}
Jawaban
x
n
+
4
n
x
n
+
16
n
n
−
1
2
=
0
x
n
+
4
n
x
n
+
16
n
n
=
1
2
x
n
+
4
n
x
n
+
16
n
=
(
1
2
)
n
x
n
+
4
n
x
n
+
16
n
=
1
2
n
2
n
(
x
n
+
4
n
)
=
x
n
+
16
n
2
n
(
x
n
+
2
2
n
)
=
x
n
+
2
4
n
2
n
⋅
x
n
+
2
3
n
=
x
n
+
2
4
n
2
n
⋅
x
n
−
x
n
=
2
4
n
−
2
3
n
x
n
(
2
n
−
1
)
=
2
3
n
(
2
n
−
1
)
x
n
=
2
3
n
x
n
=
(
2
3
)
n
x
n
=
8
n
x
=
8
{\displaystyle {\begin{aligned}{\sqrt[{n}]{\frac {x^{n}+4^{n}}{x^{n}+16^{n}}}}-{\frac {1}{2}}&=0\\{\sqrt[{n}]{\frac {x^{n}+4^{n}}{x^{n}+16^{n}}}}&={\frac {1}{2}}\\{\frac {x^{n}+4^{n}}{x^{n}+16^{n}}}&=({\frac {1}{2}})^{n}\\{\frac {x^{n}+4^{n}}{x^{n}+16^{n}}}&={\frac {1}{2^{n}}}\\2^{n}(x^{n}+4^{n})&=x^{n}+16^{n}\\2^{n}(x^{n}+2^{2n})&=x^{n}+2^{4n}\\2^{n}\cdot x^{n}+2^{3n}&=x^{n}+2^{4n}\\2^{n}\cdot x^{n}-x^{n}&=2^{4n}-2^{3n}\\x^{n}(2^{n}-1)&=2^{3n}(2^{n}-1)\\x^{n}&=2^{3n}\\x^{n}&=(2^{3})^{n}\\x^{n}&=8^{n}\\x&=8\\\end{aligned}}}
Berapa hasil dari
30
+
25
+
24
+
20
20
+
6
+
4
{\displaystyle {\frac {{\sqrt {30}}+{\sqrt {25}}+{\sqrt {24}}+{\sqrt {20}}}{{\sqrt {20}}+{\sqrt {6}}+{\sqrt {4}}}}}
Jawaban
misalkan
x
=
30
+
25
+
24
+
20
20
+
6
+
4
x
=
30
+
25
+
24
+
20
20
+
6
+
4
=
5
⋅
6
+
5
⋅
5
+
6
⋅
4
+
5
⋅
4
5
⋅
4
+
6
+
4
=
5
⋅
6
+
5
⋅
5
+
6
⋅
4
+
5
⋅
4
2
⋅
5
+
6
+
4
=
6
⋅
5
+
6
⋅
4
+
5
⋅
5
+
5
⋅
4
5
+
6
+
5
+
4
=
6
(
5
+
4
)
+
5
(
5
+
4
)
6
+
5
+
5
+
4
=
(
6
+
5
)
(
5
+
4
)
6
+
5
+
5
+
4
1
x
=
6
+
5
+
5
+
4
(
6
+
5
)
(
5
+
4
)
=
6
+
5
(
6
+
5
)
(
5
+
4
)
+
5
+
4
(
6
+
5
)
(
5
+
4
)
=
1
5
+
4
+
1
6
+
5
=
5
−
4
5
−
4
+
6
−
5
6
−
5
=
5
−
4
1
+
6
−
5
1
=
5
−
4
+
6
−
5
=
6
−
4
=
6
−
2
x
=
1
6
−
2
=
6
+
2
6
−
4
=
6
+
2
2
=
1
+
6
2
{\displaystyle {\begin{aligned}{\text{misalkan }}x={\frac {{\sqrt {30}}+{\sqrt {25}}+{\sqrt {24}}+{\sqrt {20}}}{{\sqrt {20}}+{\sqrt {6}}+{\sqrt {4}}}}\\x&={\frac {{\sqrt {30}}+{\sqrt {25}}+{\sqrt {24}}+{\sqrt {20}}}{{\sqrt {20}}+{\sqrt {6}}+{\sqrt {4}}}}\\&={\frac {{\sqrt {5\cdot 6}}+{\sqrt {5\cdot 5}}+{\sqrt {6\cdot 4}}+{\sqrt {5\cdot 4}}}{{\sqrt {5\cdot 4}}+{\sqrt {6}}+{\sqrt {4}}}}\\&={\frac {{\sqrt {5}}\cdot {\sqrt {6}}+{\sqrt {5}}\cdot {\sqrt {5}}+{\sqrt {6}}\cdot {\sqrt {4}}+{\sqrt {5}}\cdot {\sqrt {4}}}{2\cdot {\sqrt {5}}+{\sqrt {6}}+{\sqrt {4}}}}\\&={\frac {{\sqrt {6}}\cdot {\sqrt {5}}+{\sqrt {6}}\cdot {\sqrt {4}}+{\sqrt {5}}\cdot {\sqrt {5}}+{\sqrt {5}}\cdot {\sqrt {4}}}{{\sqrt {5}}+{\sqrt {6}}+{\sqrt {5}}+{\sqrt {4}}}}\\&={\frac {{\sqrt {6}}({\sqrt {5}}+{\sqrt {4}})+{\sqrt {5}}({\sqrt {5}}+{\sqrt {4}})}{{\sqrt {6}}+{\sqrt {5}}+{\sqrt {5}}+{\sqrt {4}}}}\\&={\frac {({\sqrt {6}}+{\sqrt {5}})({\sqrt {5}}+{\sqrt {4}})}{{\sqrt {6}}+{\sqrt {5}}+{\sqrt {5}}+{\sqrt {4}}}}\\{\frac {1}{x}}&={\frac {{\sqrt {6}}+{\sqrt {5}}+{\sqrt {5}}+{\sqrt {4}}}{({\sqrt {6}}+{\sqrt {5}})({\sqrt {5}}+{\sqrt {4}})}}\\&={\frac {{\sqrt {6}}+{\sqrt {5}}}{({\sqrt {6}}+{\sqrt {5}})({\sqrt {5}}+{\sqrt {4}})}}+{\frac {{\sqrt {5}}+{\sqrt {4}}}{({\sqrt {6}}+{\sqrt {5}})({\sqrt {5}}+{\sqrt {4}})}}\\&={\frac {1}{{\sqrt {5}}+{\sqrt {4}}}}+{\frac {1}{{\sqrt {6}}+{\sqrt {5}}}}\\&={\frac {{\sqrt {5}}-{\sqrt {4}}}{5-4}}+{\frac {{\sqrt {6}}-{\sqrt {5}}}{6-5}}\\&={\frac {{\sqrt {5}}-{\sqrt {4}}}{1}}+{\frac {{\sqrt {6}}-{\sqrt {5}}}{1}}\\&={\sqrt {5}}-{\sqrt {4}}+{\sqrt {6}}-{\sqrt {5}}\\&={\sqrt {6}}-{\sqrt {4}}\\&={\sqrt {6}}-2\\x&={\frac {1}{{\sqrt {6}}-2}}\\&={\frac {{\sqrt {6}}+2}{6-4}}\\&={\frac {{\sqrt {6}}+2}{2}}\\&=1+{\frac {\sqrt {6}}{2}}\\\end{aligned}}}
Berapa hasil dari
(
6
+
2
32
)
5
{\displaystyle ({\frac {{\sqrt {6}}+{\sqrt {2}}}{\sqrt {32}}})^{5}}
Jawaban
(
6
+
2
32
)
5
misalkan
x
=
6
+
2
32
x
=
6
+
2
32
=
2
(
3
+
1
)
4
2
=
3
+
1
4
4
x
=
3
+
1
4
x
−
1
=
3
(
4
x
−
1
)
2
=
3
16
x
2
−
8
x
+
1
=
3
16
x
2
=
8
x
+
2
8
x
2
=
4
x
+
1
x
2
=
4
x
+
1
8
cara 1
x
3
=
x
⋅
x
2
=
x
(
4
x
+
1
8
)
=
4
x
2
+
x
8
=
4
x
2
8
+
x
8
=
4
(
4
x
+
1
8
)
8
+
x
8
=
16
x
+
4
64
+
x
8
=
4
x
+
1
16
+
x
8
=
4
x
+
1
+
2
x
16
=
6
x
+
1
16
x
5
=
x
2
⋅
x
3
=
(
4
x
+
1
8
)
(
6
x
+
1
16
)
=
24
x
2
+
10
x
+
1
128
=
24
x
2
128
+
10
x
+
1
128
=
24
(
4
x
+
1
8
)
128
+
10
x
+
1
128
=
96
x
+
24
1024
+
10
x
+
1
128
=
96
x
+
24
+
80
x
+
8
1024
=
176
x
+
32
1024
=
176
x
1024
+
32
1024
=
176
1024
(
3
+
1
4
)
+
32
1024
=
44
(
3
+
1
)
1024
+
32
1024
=
44
3
+
44
1024
+
32
1024
=
76
+
44
3
1024
=
19
+
11
3
256
cara 2
x
4
=
(
x
2
)
2
=
(
4
x
+
1
8
)
2
=
16
x
2
+
8
x
+
1
64
=
16
x
2
64
+
8
x
64
+
1
64
=
x
2
4
+
x
8
+
1
64
=
4
x
+
1
8
4
+
x
8
+
1
64
=
4
x
32
+
1
32
+
x
8
+
1
64
=
x
8
+
1
32
+
x
8
+
1
64
=
x
4
+
3
64
x
5
=
x
⋅
x
4
=
(
3
+
1
4
)
(
x
4
+
3
64
)
=
(
3
+
1
4
)
(
3
+
1
4
4
+
3
64
)
=
(
3
+
1
4
)
(
3
+
1
16
+
3
64
)
=
(
3
+
1
)
2
64
+
(
3
+
1
4
)
3
64
=
3
+
2
3
+
1
64
+
3
(
3
+
1
)
256
=
4
+
2
3
64
+
3
(
3
+
1
)
256
=
16
+
8
3
256
+
3
3
+
3
256
=
19
+
11
3
256
{\displaystyle {\begin{aligned}({\frac {{\sqrt {6}}+{\sqrt {2}}}{\sqrt {32}}})^{5}\\{\text{misalkan }}x={\frac {{\sqrt {6}}+{\sqrt {2}}}{\sqrt {32}}}\\x&={\frac {{\sqrt {6}}+{\sqrt {2}}}{\sqrt {32}}}\\&={\frac {{\sqrt {2}}({\sqrt {3}}+1)}{4{\sqrt {2}}}}\\&={\frac {{\sqrt {3}}+1}{4}}\\4x&={\sqrt {3}}+1\\4x-1&={\sqrt {3}}\\(4x-1)^{2}&=3\\16x^{2}-8x+1&=3\\16x^{2}&=8x+2\\8x^{2}&=4x+1\\x^{2}&={\frac {4x+1}{8}}\\{\text{cara 1 }}\\x^{3}&=x\cdot x^{2}\\&=x({\frac {4x+1}{8}})\\&={\frac {4x^{2}+x}{8}}\\&={\frac {4x^{2}}{8}}+{\frac {x}{8}}\\&={\frac {4({\frac {4x+1}{8}})}{8}}+{\frac {x}{8}}\\&={\frac {16x+4}{64}}+{\frac {x}{8}}\\&={\frac {4x+1}{16}}+{\frac {x}{8}}\\&={\frac {4x+1+2x}{16}}\\&={\frac {6x+1}{16}}\\x^{5}&=x^{2}\cdot x^{3}\\&=({\frac {4x+1}{8}})({\frac {6x+1}{16}})\\&={\frac {24x^{2}+10x+1}{128}}\\&={\frac {24x^{2}}{128}}+{\frac {10x+1}{128}}\\&={\frac {24({\frac {4x+1}{8}})}{128}}+{\frac {10x+1}{128}}\\&={\frac {96x+24}{1024}}+{\frac {10x+1}{128}}\\&={\frac {96x+24+80x+8}{1024}}\\&={\frac {176x+32}{1024}}\\&={\frac {176x}{1024}}+{\frac {32}{1024}}\\&={\frac {176}{1024}}({\frac {{\sqrt {3}}+1}{4}})+{\frac {32}{1024}}\\&={\frac {44({\sqrt {3}}+1)}{1024}}+{\frac {32}{1024}}\\&={\frac {44{\sqrt {3}}+44}{1024}}+{\frac {32}{1024}}\\&={\frac {76+44{\sqrt {3}}}{1024}}\\&={\frac {19+11{\sqrt {3}}}{256}}\\{\text{cara 2 }}\\x^{4}&=(x^{2})^{2}\\&=({\frac {4x+1}{8}})^{2}\\&={\frac {16x^{2}+8x+1}{64}}\\&={\frac {16x^{2}}{64}}+{\frac {8x}{64}}+{\frac {1}{64}}\\&={\frac {x^{2}}{4}}+{\frac {x}{8}}+{\frac {1}{64}}\\&={\frac {\frac {4x+1}{8}}{4}}+{\frac {x}{8}}+{\frac {1}{64}}\\&={\frac {4x}{32}}+{\frac {1}{32}}+{\frac {x}{8}}+{\frac {1}{64}}\\&={\frac {x}{8}}+{\frac {1}{32}}+{\frac {x}{8}}+{\frac {1}{64}}\\&={\frac {x}{4}}+{\frac {3}{64}}\\x^{5}&=x\cdot x^{4}\\&=({\frac {{\sqrt {3}}+1}{4}})({\frac {x}{4}}+{\frac {3}{64}})\\&=({\frac {{\sqrt {3}}+1}{4}})({\frac {\frac {{\sqrt {3}}+1}{4}}{4}}+{\frac {3}{64}})\\&=({\frac {{\sqrt {3}}+1}{4}})({\frac {{\sqrt {3}}+1}{16}}+{\frac {3}{64}})\\&={\frac {({\sqrt {3}}+1)^{2}}{64}}+({\frac {{\sqrt {3}}+1}{4}}){\frac {3}{64}}\\&={\frac {3+2{\sqrt {3}}+1}{64}}+{\frac {3({\sqrt {3}}+1)}{256}}\\&={\frac {4+2{\sqrt {3}}}{64}}+{\frac {3({\sqrt {3}}+1)}{256}}\\&={\frac {16+8{\sqrt {3}}}{256}}+{\frac {3{\sqrt {3}}+3}{256}}\\&={\frac {19+11{\sqrt {3}}}{256}}\\\end{aligned}}}
Berapa hasil dari
1
4
+
5
16
+
9
64
+
13
256
+
…
{\displaystyle {\frac {1}{4}}+{\frac {5}{16}}+{\frac {9}{64}}+{\frac {13}{256}}+\dots }
Jawaban
x
=
1
4
+
5
16
+
9
64
+
13
256
+
…
x
4
=
1
16
+
5
64
+
9
256
+
13
1.024
+
…
3
x
4
=
1
4
+
4
16
+
4
64
+
4
256
+
…
3
x
4
=
1
4
+
4
(
1
16
+
1
64
+
1
256
+
…
)
1
16
+
1
64
+
1
256
+
…
=
1
1
−
1
4
=
4
3
3
x
4
=
1
4
+
4
(
4
3
)
=
1
4
+
16
3
=
67
12
x
=
67
9
{\displaystyle {\begin{aligned}x&={\frac {1}{4}}+{\frac {5}{16}}+{\frac {9}{64}}+{\frac {13}{256}}+\dots \\{\frac {x}{4}}&={\frac {1}{16}}+{\frac {5}{64}}+{\frac {9}{256}}+{\frac {13}{1.024}}+\dots \\{\frac {3x}{4}}&={\frac {1}{4}}+{\frac {4}{16}}+{\frac {4}{64}}+{\frac {4}{256}}+\dots \\{\frac {3x}{4}}&={\frac {1}{4}}+4({\frac {1}{16}}+{\frac {1}{64}}+{\frac {1}{256}}+\dots )\\{\frac {1}{16}}+{\frac {1}{64}}+{\frac {1}{256}}+\dots &={\frac {1}{1-{\frac {1}{4}}}}\\&={\frac {4}{3}}\\{\frac {3x}{4}}&={\frac {1}{4}}+4({\frac {4}{3}})\\&={\frac {1}{4}}+{\frac {16}{3}}\\&={\frac {67}{12}}\\x&={\frac {67}{9}}\\\end{aligned}}}
Berapa nilai y-x jika
1
+
2
+
3
+
4
+
⋯
+
106
4
+
5
+
6
+
7
+
⋯
+
109
=
x
y
{\displaystyle {\frac {1+2+3+4+\dots +106}{4+5+6+7+\dots +109}}={\frac {x}{y}}}
Jawaban
1
+
2
+
3
+
4
+
⋯
+
106
4
+
5
+
6
+
7
+
⋯
+
109
=
x
y
106
×
107
2
106
2
(
4
+
109
)
=
x
y
53
×
107
53
×
113
=
x
y
y
−
x
=
113
−
107
=
6
{\displaystyle {\begin{aligned}{\frac {1+2+3+4+\dots +106}{4+5+6+7+\dots +109}}&={\frac {x}{y}}\\{\frac {\frac {106\times 107}{2}}{{\frac {106}{2}}(4+109)}}&={\frac {x}{y}}\\{\frac {53\times 107}{53\times 113}}&={\frac {x}{y}}\\y-x&=113-107=6\\\end{aligned}}}
Berapa angka satuan dari hasil 172024 ? Jawaban
Perhatikan angka satuannya
17
1
=
7
17
2
=
9
17
3
=
3
17
4
=
1
17
5
=
7
17
6
=
9
17
7
=
3
17
8
=
1
Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1
{\displaystyle {\begin{aligned}{\text{Perhatikan angka satuannya}}\\17^{1}&=7\\17^{2}&=9\\17^{3}&=3\\17^{4}&=1\\17^{5}&=7\\17^{6}&=9\\17^{7}&=3\\17^{8}&=1\\{\text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1}}\end{aligned}}}
Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!? Jawaban
Perhatikan
1
!
+
2
!
+
3
!
+
4
!
+
⋯
+
2024
!
=
1
+
(
1
x
2
)
+
(
1
x
2
x
3
)
+
(
1
x
2
x
3
x
4
)
+
⋯
+
2024
!
=
1
+
2
+
6
+
24
+
120
+
720
+
⋯
+
2024
!
Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka
1
+
2
+
6
+
24
=
33
jadi angka satuannya adalah
3
{\displaystyle {\begin{aligned}{\text{Perhatikan}}\\1!+2!+3!+4!+\dots +2024!&=1+(1x2)+(1x2x3)+(1x2x3x4)+\dots +2024!\\&=1+2+6+24+120+720+\dots +2024!\\{\text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka }}1+2+6+24=33{\text{ jadi angka satuannya adalah }}3\end{aligned}}}
Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12? Jawaban
Perhatikan
1
!
+
2
!
+
3
!
+
4
!
+
⋯
+
2024
!
12
=
1
+
1
x
2
+
1
x
2
x
3
+
1
x
2
x
3
x
4
+
⋯
+
2024
!
12
=
1
+
2
+
6
+
24
+
⋯
+
2024
!
12
karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi
1
+
2
+
6
=
9
{\displaystyle {\begin{aligned}{\text{Perhatikan}}\\{\frac {1!+2!+3!+4!+\dots +2024!}{12}}&={\frac {1+1x2+1x2x3+1x2x3x4+\dots +2024!}{12}}\\&={\frac {1+2+6+24+\dots +2024!}{12}}\\{\text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi }}1+2+6=9\end{aligned}}}
Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A? Jawaban
penjumlahan
1
+
1
+
1
+
1
+
…
(sebanyak 88 buah)
+
x
+
y
=
A
88
+
x
+
y
=
A
perkalian
1
×
1
×
1
×
…
(sebanyak 88 buah)
×
x
×
y
=
A
x
×
y
=
A
88
+
x
+
y
=
x
y
x
y
−
y
=
88
+
x
y
(
x
−
1
)
=
88
+
x
y
=
88
+
x
x
−
1
uji selidiki untuk x=2
y
=
88
+
2
2
−
1
=
90
buktikan
88
+
x
+
y
=
x
y
88
+
2
+
90
=
2
(
90
)
180
=
180
terbukti
nilai A adalah
180
{\displaystyle {\begin{aligned}{\text{penjumlahan}}\\1+1+1+1+\dots {\text{ (sebanyak 88 buah) }}+x+y&=A\\88+x+y&=A\\{\text{perkalian}}\\1\times 1\times 1\times \dots {\text{ (sebanyak 88 buah) }}\times x\times y&=A\\x\times y&=A\\88+x+y&=xy\\xy-y&=88+x\\y(x-1)&=88+x\\y&={\frac {88+x}{x-1}}\\{\text{uji selidiki untuk x=2}}\\y&={\frac {88+2}{2-1}}\\&=90\\{\text{buktikan}}\\88+x+y&=xy\\88+2+90&=2(90)\\180&=180\\{\text{terbukti}}\\{\text{nilai A adalah }}180\\\end{aligned}}}
Berapakah nilai x, y dan z dari
x
+
y
−
z
=
1
,
x
2
+
y
2
−
z
2
=
−
5
dan
x
3
+
y
3
−
z
3
=
−
53
{\displaystyle x+y-z=1,x^{2}+y^{2}-z^{2}=-5{\text{ dan }}x^{3}+y^{3}-z^{3}=-53}
Jawaban
x
+
y
−
z
=
1
x
+
y
=
z
+
1
x
2
+
2
x
y
+
y
2
=
z
2
+
2
z
+
1
x
2
+
y
2
−
z
2
=
2
z
+
1
−
2
x
y
−
5
=
2
z
+
1
−
2
x
y
2
x
y
=
2
z
+
6
x
y
=
z
+
3
x
2
+
y
2
−
z
2
=
−
5
x
2
+
y
2
=
z
2
−
5
x
3
+
y
3
−
z
3
=
−
53
(
x
+
y
)
(
x
2
−
x
y
+
y
2
)
−
z
3
+
53
=
0
(
x
+
y
)
(
x
2
+
y
2
−
x
y
)
−
z
3
+
53
=
0
(
z
+
1
)
(
z
2
−
5
−
(
z
+
3
)
)
−
z
3
+
53
=
0
(
z
+
1
)
(
z
2
−
z
−
8
)
−
z
3
+
53
=
0
z
3
−
z
2
−
8
z
+
z
2
−
z
−
8
−
z
3
+
53
=
0
−
9
z
+
45
=
0
−
9
z
=
−
45
z
=
5
x
+
y
=
5
+
1
x
+
y
=
6
x
=
6
−
y
x
y
=
5
+
3
x
y
=
8
(
6
−
y
)
y
=
8
6
y
−
y
2
=
8
y
2
−
6
y
+
8
=
0
(
y
−
4
)
(
y
−
2
)
=
0
y
=
4
atau
y
=
2
jika
y
=
4
x
+
y
=
z
+
1
x
+
4
=
5
+
1
x
=
2
jika
y
=
2
x
+
y
=
z
+
1
x
+
2
=
5
+
1
x
=
4
{\displaystyle {\begin{aligned}x+y-z&=1\\x+y&=z+1\\x^{2}+2xy+y^{2}&=z^{2}+2z+1\\x^{2}+y^{2}-z^{2}&=2z+1-2xy\\-5&=2z+1-2xy\\2xy&=2z+6\\xy&=z+3\\x^{2}+y^{2}-z^{2}&=-5\\x^{2}+y^{2}&=z^{2}-5\\x^{3}+y^{3}-z^{3}&=-53\\(x+y)(x^{2}-xy+y^{2})-z^{3}+53&=0\\(x+y)(x^{2}+y^{2}-xy)-z^{3}+53&=0\\(z+1)(z^{2}-5-(z+3))-z^{3}+53&=0\\(z+1)(z^{2}-z-8)-z^{3}+53&=0\\z^{3}-z^{2}-8z+z^{2}-z-8-z^{3}+53&=0\\-9z+45&=0\\-9z&=-45\\z&=5\\x+y&=5+1\\x+y&=6\\x&=6-y\\xy&=5+3\\xy&=8\\(6-y)y&=8\\6y-y^{2}&=8\\y^{2}-6y+8&=0\\(y-4)(y-2)&=0\\y=4{\text{ atau }}y=2\\{\text{jika }}y=4\\x+y&=z+1\\x+4&=5+1\\x&=2\\{\text{jika }}y=2\\x+y&=z+1\\x+2&=5+1\\x&=4\\\end{aligned}}}
Berapakah nilai titik koordinat (x,y) dari
x
+
y
+
x
−
y
=
432
x
13
y
{\displaystyle {\sqrt {x+y}}+{\sqrt {x-y}}={\sqrt {\frac {432x}{13y}}}}
x
+
y
−
x
−
y
=
52
y
3
x
{\displaystyle {\sqrt {x+y}}-{\sqrt {x-y}}={\sqrt {\frac {52y}{3x}}}}
Jawaban
x
+
y
+
x
−
y
=
432
x
13
y
x
+
y
−
x
−
y
=
52
y
3
x
(
x
+
y
+
x
−
y
)
(
x
+
y
−
x
−
y
)
=
432
x
13
y
⋅
52
y
3
x
x
+
y
−
x
+
y
=
432
x
⋅
52
y
13
y
⋅
3
x
2
y
=
144
⋅
4
2
y
=
576
2
y
=
24
y
=
12
x
+
12
+
x
−
12
=
432
x
13
y
x
+
12
+
x
−
12
=
432
x
13
(
12
)
x
+
12
+
x
−
12
+
2
⋅
x
+
12
⋅
x
−
12
=
36
x
13
2
x
+
2
x
2
−
144
=
36
x
13
2
(
x
+
x
2
−
144
)
=
36
x
13
x
+
x
2
−
144
=
18
x
13
x
2
−
144
=
5
x
13
x
2
−
144
=
25
x
2
169
144
x
2
169
−
144
=
0
x
2
169
−
1
=
0
x
2
−
169
=
0
(
x
−
13
)
(
x
+
13
)
=
0
x
1
=
13
atau
x
2
=
−
13
(TM) karena
x
>
y
{\displaystyle {\begin{aligned}{\sqrt {x+y}}+{\sqrt {x-y}}&={\sqrt {\frac {432x}{13y}}}\\{\sqrt {x+y}}-{\sqrt {x-y}}&={\sqrt {\frac {52y}{3x}}}\\({\sqrt {x+y}}+{\sqrt {x-y}})({\sqrt {x+y}}-{\sqrt {x-y}})&={\sqrt {\frac {432x}{13y}}}\cdot {\sqrt {\frac {52y}{3x}}}\\x+y-x+y&={\sqrt {\frac {432x\cdot 52y}{13y\cdot 3x}}}\\2y&={\sqrt {144\cdot 4}}\\2y&={\sqrt {576}}\\2y&=24\\y&=12\\{\sqrt {x+12}}+{\sqrt {x-12}}&={\sqrt {\frac {432x}{13y}}}\\{\sqrt {x+12}}+{\sqrt {x-12}}&={\sqrt {\frac {432x}{13(12)}}}\\x+12+x-12+2\cdot {\sqrt {x+12}}\cdot {\sqrt {x-12}}&={\frac {36x}{13}}\\2x+2{\sqrt {x^{2}-144}}&={\frac {36x}{13}}\\2(x+{\sqrt {x^{2}-144}})&={\frac {36x}{13}}\\x+{\sqrt {x^{2}-144}}&={\frac {18x}{13}}\\{\sqrt {x^{2}-144}}&={\frac {5x}{13}}\\x^{2}-144&={\frac {25x^{2}}{169}}\\{\frac {144x^{2}}{169}}-144&=0\\{\frac {x^{2}}{169}}-1&=0\\x^{2}-169&=0\\(x-13)(x+13)&=0\\x_{1}=13&{\text{ atau }}x_{2}=-13{\text{ (TM) karena }}x>y\\\end{aligned}}}
jadi titik koordinat (13,12)
Berapakah nilai dari x2 -7x jika
(
x
−
2
)
2
+
1
(
x
−
2
)
2
=
11
{\displaystyle (x-2)^{2}+{\frac {1}{(x-2)^{2}}}=11}
Jawaban
(
x
−
2
)
2
+
1
(
x
−
2
)
2
=
11
(
x
−
2
)
2
−
2
(
x
−
2
)
1
(
x
−
2
)
+
1
(
x
−
2
)
2
=
11
−
2
(
x
−
2
−
1
x
−
2
)
2
=
9
x
−
2
−
1
x
−
2
=
3
(
x
−
2
)
2
−
1
=
3
(
x
−
2
)
x
2
−
4
x
+
4
−
1
=
3
x
−
6
x
2
−
7
x
=
−
9
{\displaystyle {\begin{aligned}(x-2)^{2}+{\frac {1}{(x-2)^{2}}}&=11\\(x-2)^{2}-2(x-2){\frac {1}{(x-2)}}+{\frac {1}{(x-2)^{2}}}&=11-2\\(x-2-{\frac {1}{x-2}})^{2}&=9\\x-2-{\frac {1}{x-2}}&=3\\(x-2)^{2}-1&=3(x-2)\\x^{2}-4x+4-1&=3x-6\\x^{2}-7x&=-9\\\end{aligned}}}
Berapakah nilai dari
20
x
y
z
x
y
+
y
z
+
x
z
{\displaystyle {\frac {20xyz}{xy+yz+xz}}}
16
x
=
256
y
=
625
z
=
40
{\displaystyle 16^{x}=256^{y}=625^{z}=40}
Jawaban
16
x
=
256
y
=
625
z
=
40
2
4
x
=
4
4
y
=
5
4
z
=
40
2
4
x
=
40
2
=
40
1
4
x
4
4
y
=
40
4
=
40
1
4
y
5
4
z
=
40
5
=
40
1
4
z
2
⋅
4
⋅
5
=
40
1
4
x
⋅
40
1
4
y
⋅
40
1
4
z
40
=
40
1
4
x
⋅
40
1
4
y
⋅
40
1
4
z
40
=
40
1
4
x
+
1
4
y
+
1
4
z
1
=
1
4
x
+
1
4
y
+
1
4
z
4
=
1
x
+
1
y
+
1
z
20
x
y
z
x
y
+
y
z
+
x
z
=
20
⋅
x
y
z
x
y
+
y
z
+
x
z
=
20
⋅
(
x
y
+
y
z
+
x
z
x
y
z
)
−
1
=
20
⋅
(
1
z
+
1
x
+
1
y
)
−
1
=
20
⋅
(
1
x
+
1
y
+
1
z
)
−
1
=
20
⋅
(
4
)
−
1
=
20
⋅
1
4
=
5
{\displaystyle {\begin{aligned}16^{x}=256^{y}=625^{z}&=40\\2^{4x}=4^{4y}=5^{4z}&=40\\2^{4x}&=40\\2&=40^{\frac {1}{4x}}\\4^{4y}&=40\\4&=40^{\frac {1}{4y}}\\5^{4z}&=40\\5&=40^{\frac {1}{4z}}\\2\cdot 4\cdot 5&=40^{\frac {1}{4x}}\cdot 40^{\frac {1}{4y}}\cdot 40^{\frac {1}{4z}}\\40&=40^{\frac {1}{4x}}\cdot 40^{\frac {1}{4y}}\cdot 40^{\frac {1}{4z}}\\40&=40^{{\frac {1}{4x}}+{\frac {1}{4y}}+{\frac {1}{4z}}}\\1&={\frac {1}{4x}}+{\frac {1}{4y}}+{\frac {1}{4z}}\\4&={\frac {1}{x}}+{\frac {1}{y}}+{\frac {1}{z}}\\{\frac {20xyz}{xy+yz+xz}}&=20\cdot {\frac {xyz}{xy+yz+xz}}\\&=20\cdot ({\frac {xy+yz+xz}{xyz}})^{-1}\\&=20\cdot ({\frac {1}{z}}+{\frac {1}{x}}+{\frac {1}{y}})^{-1}\\&=20\cdot ({\frac {1}{x}}+{\frac {1}{y}}+{\frac {1}{z}})^{-1}\\&=20\cdot (4)^{-1}\\&=20\cdot {\frac {1}{4}}\\&=5\\\end{aligned}}}
Berapakah nilai dari
x
2
x
4
+
3
x
2
+
1
{\displaystyle {\frac {x^{2}}{x^{4}+3x^{2}+1}}}
6
x
2
+
25
x
+
6
=
0
{\displaystyle 6x^{2}+25x+6=0}
Jawaban
6
x
2
+
25
x
+
6
=
0
6
x
+
25
+
6
x
=
0
6
(
x
+
1
x
)
=
−
25
x
+
1
x
=
−
25
6
(
c
+
1
x
)
2
=
(
−
25
6
)
2
x
2
+
2
+
1
x
2
=
625
36
x
2
+
1
x
2
=
625
36
−
2
x
2
+
1
x
2
=
553
36
x
2
x
4
+
3
x
2
+
1
=
1
x
2
+
3
+
1
x
2
=
1
a
2
+
1
x
2
+
3
=
1
553
36
+
3
=
1
661
36
=
36
661
{\displaystyle {\begin{aligned}6x^{2}+25x+6&=0\\6x+25+{\frac {6}{x}}&=0\\6(x+{\frac {1}{x}})&=-25\\x+{\frac {1}{x}}&={\frac {-25}{6}}\\(c+{\frac {1}{x}})^{2}&=({\frac {-25}{6}})^{2}\\x^{2}+2+{\frac {1}{x^{2}}}&={\frac {625}{36}}\\x^{2}+{\frac {1}{x^{2}}}&={\frac {625}{36}}-2\\x^{2}+{\frac {1}{x^{2}}}&={\frac {553}{36}}\\{\frac {x^{2}}{x^{4}+3x^{2}+1}}&={\frac {1}{x^{2}+3+{\frac {1}{x^{2}}}}}\\&={\frac {1}{a^{2}+{\frac {1}{x^{2}}}+3}}\\&={\frac {1}{{\frac {553}{36}}+3}}\\&={\frac {1}{\frac {661}{36}}}\\&={\frac {36}{661}}\\\end{aligned}}}
Berapakah nilai dari
(
9
+
4
5
)
1013
(
38
+
17
5
)
675
+
6
−
5
{\displaystyle {\frac {(9+4{\sqrt {5}})^{1013}}{(38+17{\sqrt {5}})^{675}}}+6-{\sqrt {5}}}
Jawaban
(
9
+
4
5
)
1013
(
38
+
17
5
)
675
+
6
−
5
=
(
9
+
2
20
)
1013
(
(
2
)
3
+
3
(
2
)
2
(
5
)
+
3
(
2
)
(
5
)
2
+
(
5
)
3
)
675
+
6
−
5
=
(
(
2
+
5
)
2
)
1013
(
(
2
+
5
)
3
)
675
+
6
−
5
=
(
2
+
5
)
2026
(
2
+
5
)
2025
+
6
−
5
=
2
+
5
+
6
−
5
=
8
{\displaystyle {\begin{aligned}{\frac {(9+4{\sqrt {5}})^{1013}}{(38+17{\sqrt {5}})^{675}}}+6-{\sqrt {5}}&={\frac {(9+2{\sqrt {20}})^{1013}}{((2)^{3}+3(2)^{2}({\sqrt {5}})+3(2)({\sqrt {5}})^{2}+({\sqrt {5}})^{3})^{675}}}+6-{\sqrt {5}}\\&={\frac {((2+{\sqrt {5}})^{2})^{1013}}{((2+{\sqrt {5}})^{3})^{675}}}+6-{\sqrt {5}}\\&={\frac {(2+{\sqrt {5}})^{2026}}{(2+{\sqrt {5}})^{2025}}}+6-{\sqrt {5}}\\&=2+{\sqrt {5}}+6-{\sqrt {5}}\\&=8\\\end{aligned}}}
Berapakah nilai dari
27
x
3
+
8
x
3
{\displaystyle 27x^{3}+{\frac {8}{x^{3}}}}
3
x
+
2
x
=
6
{\displaystyle 3x+{\frac {2}{x}}=6}
Jawaban
3
x
+
2
x
=
6
(
3
x
+
2
x
)
3
=
6
3
27
x
3
+
3
(
3
x
)
(
2
x
)
(
3
x
+
2
x
)
+
8
x
3
=
216
27
x
3
+
18
(
6
)
+
8
x
3
=
216
27
x
3
+
108
+
8
x
3
=
216
27
x
3
+
8
x
3
=
108
{\displaystyle {\begin{aligned}3x+{\frac {2}{x}}&=6\\(3x+{\frac {2}{x}})^{3}&=6^{3}\\27x^{3}+3(3x)({\frac {2}{x}})(3x+{\frac {2}{x}})+{\frac {8}{x^{3}}}&=216\\27x^{3}+18(6)+{\frac {8}{x^{3}}}&=216\\27x^{3}+108+{\frac {8}{x^{3}}}&=216\\27x^{3}+{\frac {8}{x^{3}}}&=108\\\end{aligned}}}
Berapakah nilai dari
x
6
+
8
x
3
{\displaystyle x^{6}+{\frac {8}{x^{3}}}}
x
3
+
1
x
3
=
8
{\displaystyle x^{3}+{\frac {1}{x^{3}}}=8}
Jawaban
x
3
+
1
x
3
=
8
x
3
=
8
−
1
x
3
x
6
=
8
x
3
−
1
x
6
+
8
x
3
=
8
x
3
−
1
+
8
x
3
=
8
x
3
+
8
x
3
−
1
=
8
(
x
3
+
1
x
3
)
−
1
=
8
(
8
)
−
1
=
63
{\displaystyle {\begin{aligned}x^{3}+{\frac {1}{x^{3}}}&=8\\x^{3}&=8-{\frac {1}{x^{3}}}\\x^{6}&=8x^{3}-1\\x^{6}+{\frac {8}{x^{3}}}&=8x^{3}-1+{\frac {8}{x^{3}}}\\&=8x^{3}+{\frac {8}{x^{3}}}-1\\&=8(x^{3}+{\frac {1}{x^{3}}})-1\\&=8(8)-1\\&=63\\\end{aligned}}}
Berapakah nilai dari
4
x
+
25
x
{\displaystyle 4x+{\frac {25}{x}}}
2
x
+
5
x
=
4
x
−
25
x
{\displaystyle 2{\sqrt {x}}+{\frac {5}{\sqrt {x}}}=4x-{\frac {25}{x}}}
Jawaban
2
x
+
5
x
=
4
x
−
25
x
2
x
+
5
x
=
(
2
x
+
5
x
)
(
2
x
−
5
x
)
1
=
2
x
−
5
x
1
2
=
(
2
x
−
5
x
)
2
1
=
4
x
−
20
+
25
x
4
x
+
25
x
=
21
{\displaystyle {\begin{aligned}2{\sqrt {x}}+{\frac {5}{\sqrt {x}}}&=4x-{\frac {25}{x}}\\2{\sqrt {x}}+{\frac {5}{\sqrt {x}}}&=(2{\sqrt {x}}+{\frac {5}{\sqrt {x}}})(2{\sqrt {x}}-{\frac {5}{\sqrt {x}}})\\1&=2{\sqrt {x}}-{\frac {5}{\sqrt {x}}}\\1^{2}&=(2{\sqrt {x}}-{\frac {5}{\sqrt {x}}})^{2}\\1&=4x-20+{\frac {25}{x}}\\4x+{\frac {25}{x}}&=21\\\end{aligned}}}
Berapakah nilai dari
x
+
1
x
{\displaystyle x+{\frac {1}{x}}}
x
2
−
x
+
1
x
2
+
x
+
1
=
5
6
{\displaystyle {\frac {x^{2}-x+1}{x^{2}+x+1}}={\frac {5}{6}}}
Jawaban
x
2
−
x
+
1
x
2
+
x
+
1
=
5
6
x
2
+
1
−
x
x
2
+
1
+
x
=
5
6
x
+
1
x
−
1
x
+
1
x
+
1
=
5
6
misalkan
x
+
1
x
=
y
y
−
1
y
+
1
=
5
6
6
(
y
−
1
)
=
5
(
y
+
1
)
6
y
−
6
=
5
y
+
5
y
=
11
x
+
1
x
=
11
{\displaystyle {\begin{aligned}{\frac {x^{2}-x+1}{x^{2}+x+1}}&={\frac {5}{6}}\\{\frac {x^{2}+1-x}{x^{2}+1+x}}&={\frac {5}{6}}\\{\frac {x+{\frac {1}{x}}-1}{x+{\frac {1}{x}}+1}}&={\frac {5}{6}}\\{\text{ misalkan }}x+{\frac {1}{x}}&=y\\{\frac {y-1}{y+1}}&={\frac {5}{6}}\\6(y-1)&=5(y+1)\\6y-6&=5y+5\\y&=11\\x+{\frac {1}{x}}&=11\\\end{aligned}}}
Berapakah nilai dari
x
+
1
x
{\displaystyle x+{\frac {1}{x}}}
x
+
x
=
1
{\displaystyle {\sqrt {x}}+x=1}
Jawaban
x
+
x
=
1
x
−
1
=
−
x
(
x
−
1
)
2
=
(
−
x
)
2
x
2
−
2
x
+
1
=
x
x
2
−
3
x
+
1
=
0
x
−
3
+
1
x
=
0
x
+
1
x
=
3
{\displaystyle {\begin{aligned}{\sqrt {x}}+x&=1\\x-1&=-{\sqrt {x}}\\(x-1)^{2}&=(-{\sqrt {x}})^{2}\\x^{2}-2x+1&=x\\x^{2}-3x+1&=0\\x-3+{\frac {1}{x}}&=0\\x+{\frac {1}{x}}&=3\\\end{aligned}}}
Berapakah nilai dari
x
+
1
x
{\displaystyle x+{\frac {1}{x}}}
x
3
−
x
−
36
3
=
3
{\displaystyle {\sqrt[{3}]{x}}-{\sqrt[{3}]{x-36}}=3}
Jawaban
x
3
−
x
−
36
3
=
3
(
x
3
−
x
−
36
3
)
3
=
3
3
x
−
(
x
−
36
)
−
3
x
(
x
−
36
)
3
(
x
3
−
x
−
36
3
)
=
27
36
−
3
x
(
x
−
36
)
3
3
=
27
−
9
x
(
x
−
36
)
3
=
−
9
x
(
x
−
36
)
3
=
1
x
(
x
−
36
)
=
1
x
2
−
36
x
−
1
=
0
x
−
36
−
1
x
=
0
x
−
1
x
=
36
{\displaystyle {\begin{aligned}{\sqrt[{3}]{x}}-{\sqrt[{3}]{x-36}}&=3\\({\sqrt[{3}]{x}}-{\sqrt[{3}]{x-36}})^{3}&=3^{3}\\x-(x-36)-3{\sqrt[{3}]{x(x-36)}}({\sqrt[{3}]{x}}-{\sqrt[{3}]{x-36}})&=27\\36-3{\sqrt[{3}]{x(x-36)}}3&=27\\-9{\sqrt[{3}]{x(x-36)}}&=-9\\{\sqrt[{3}]{x(x-36)}}&=1\\x(x-36)&=1\\x^{2}-36x-1&=0\\x-36-{\frac {1}{x}}&=0\\x-{\frac {1}{x}}&=36\\\end{aligned}}}
Berapakah nilai dari
x
+
16
x
{\displaystyle x+{\frac {16}{x}}}
x
−
3
x
=
4
{\displaystyle x-3{\sqrt {x}}=4}
Jawaban
x
−
3
x
=
4
x
−
4
=
3
x
x
2
−
8
x
+
16
=
9
x
x
2
−
17
x
+
16
=
0
x
−
17
+
16
x
=
0
x
+
16
x
=
17
{\displaystyle {\begin{aligned}x-3{\sqrt {x}}&=4\\x-4&=3{\sqrt {x}}\\x^{2}-8x+16&=9x\\x^{2}-17x+16&=0\\x-17+{\frac {16}{x}}&=0\\x+{\frac {16}{x}}&=17\\\end{aligned}}}
Berapakah nilai dari
x
2
x
4
+
4
{\displaystyle {\frac {x^{2}}{x^{4}+4}}}
x
2
−
7
x
+
2
=
0
{\displaystyle x^{2}-7x+2=0}
Jawaban
x
2
−
7
x
+
2
=
0
x
2
+
2
=
7
x
x
+
2
x
=
7
x
2
+
4
+
4
x
2
=
49
x
2
+
4
x
2
=
45
x
4
+
4
x
2
=
45
x
2
x
4
+
4
=
1
45
{\displaystyle {\begin{aligned}x^{2}-7x+2&=0\\x^{2}+2&=7x\\x+{\frac {2}{x}}&=7\\x^{2}+4+{\frac {4}{x^{2}}}&=49\\x^{2}+{\frac {4}{x^{2}}}&=45\\{\frac {x^{4}+4}{x^{2}}}&=45\\{\frac {x^{2}}{x^{4}+4}}&={\frac {1}{45}}\\\end{aligned}}}
Berapakah nilai dari
x
+
x
3
4
+
x
−
3
4
+
x
−
1
{\displaystyle x+x^{\frac {3}{4}}+x^{-{\frac {3}{4}}}+x^{-1}}
x
1
4
+
x
−
1
4
=
5
{\displaystyle x^{\frac {1}{4}}+x^{-{\frac {1}{4}}}=5}
Jawaban
x
1
4
+
x
−
1
4
=
5
x
1
2
+
2
+
x
−
1
2
=
25
x
1
2
+
x
−
1
2
=
23
x
+
2
+
x
−
1
=
529
x
+
x
−
1
=
527
x
1
4
+
x
−
1
4
=
5
x
3
4
+
3
(
x
1
4
+
x
−
1
4
)
+
x
−
3
4
=
125
x
3
4
+
3
(
5
)
+
x
−
3
4
=
125
x
3
4
+
x
−
3
4
=
110
x
+
x
3
4
+
x
−
3
4
+
x
−
1
=
x
+
x
−
1
+
x
3
4
+
x
−
3
4
=
527
+
110
=
637
{\displaystyle {\begin{aligned}x^{\frac {1}{4}}+x^{-{\frac {1}{4}}}&=5\\x^{\frac {1}{2}}+2+x^{-{\frac {1}{2}}}&=25\\x^{\frac {1}{2}}+x^{-{\frac {1}{2}}}&=23\\x+2+x^{-1}&=529\\x+x^{-1}&=527\\x^{\frac {1}{4}}+x^{-{\frac {1}{4}}}&=5\\x^{\frac {3}{4}}+3(x^{\frac {1}{4}}+x^{-{\frac {1}{4}}})+x^{-{\frac {3}{4}}}&=125\\x^{\frac {3}{4}}+3(5)+x^{-{\frac {3}{4}}}&=125\\x^{\frac {3}{4}}+x^{-{\frac {3}{4}}}&=110\\x+x^{\frac {3}{4}}+x^{-{\frac {3}{4}}}+x^{-1}&=x+x^{-1}+x^{\frac {3}{4}}+x^{-{\frac {3}{4}}}\\&=527+110\\&=637\\\end{aligned}}}
Berapakah nilai dari
8
x
6
+
x
5
+
x
4
+
5
x
3
+
1
{\displaystyle {\sqrt {8x^{6}+x^{5}+x^{4}+5x^{3}+1}}}
1
x
3
+
1
x
4
+
1
x
5
=
0
{\displaystyle {\frac {1}{x^{3}}}+{\frac {1}{x^{4}}}+{\frac {1}{x^{5}}}=0}
Jawaban
1
x
3
+
1
x
4
+
1
x
5
=
0
x
2
+
x
+
1
x
5
=
0
x
2
+
x
+
1
=
0
x
2
+
x
+
1
=
0
(
x
−
1
)
(
x
2
+
x
+
1
)
=
0
(
x
−
1
)
x
3
−
1
=
0
x
3
=
1
x
=
1
8
x
6
+
x
5
+
x
4
+
5
x
3
+
1
=
(
2
x
3
)
2
+
x
3
x
2
+
x
3
x
+
5
x
3
+
1
=
(
2
(
1
)
)
2
+
(
1
)
x
2
+
(
1
)
x
+
5
(
1
)
+
1
=
(
2
)
2
+
x
2
+
x
+
5
+
1
=
4
+
x
2
+
x
+
1
+
5
=
4
+
0
+
5
=
9
=
3
{\displaystyle {\begin{aligned}{\frac {1}{x^{3}}}+{\frac {1}{x^{4}}}+{\frac {1}{x^{5}}}&=0\\{\frac {x^{2}+x+1}{x^{5}}}&=0\\x^{2}+x+1&=0\\x^{2}+x+1&=0\\(x-1)(x^{2}+x+1)&=0(x-1)\\x^{3}-1&=0\\x^{3}&=1\\x&=1\\{\sqrt {8x^{6}+x^{5}+x^{4}+5x^{3}+1}}&={\sqrt {(2x^{3})^{2}+x^{3}x^{2}+x^{3}x+5x^{3}+1}}\\&={\sqrt {(2(1))^{2}+(1)x^{2}+(1)x+5(1)+1}}\\&={\sqrt {(2)^{2}+x^{2}+x+5+1}}\\&={\sqrt {4+x^{2}+x+1+5}}\\&={\sqrt {4+0+5}}\\&={\sqrt {9}}\\&=3\\\end{aligned}}}
Berapakah nilai dari
f
(
1
)
+
f
(
2
)
+
f
(
3
)
+
⋯
+
f
(
99
)
{\displaystyle f(1)+f(2)+f(3)+\dots +f(99)}
f
(
x
)
=
1
x
+
1
+
x
{\displaystyle f(x)={\frac {1}{{\sqrt {x+1}}+{\sqrt {x}}}}}
Jawaban
f
(
x
)
=
1
x
+
1
+
x
=
x
+
1
−
x
x
+
1
−
x
=
x
+
1
−
x
f
(
1
)
+
f
(
2
)
+
f
(
3
)
+
⋯
+
f
(
98
)
+
f
(
99
)
=
1
+
1
−
1
+
2
+
1
−
2
+
3
+
1
−
3
+
⋅
+
98
+
1
−
98
+
99
+
1
−
99
=
2
−
1
+
3
−
2
+
4
−
3
+
⋅
+
99
−
98
+
100
−
99
=
100
−
1
=
10
−
1
=
9
{\displaystyle {\begin{aligned}f(x)&={\frac {1}{{\sqrt {x+1}}+{\sqrt {x}}}}\\&={\frac {{\sqrt {x+1}}-{\sqrt {x}}}{x+1-x}}\\&={\sqrt {x+1}}-{\sqrt {x}}\\f(1)+f(2)+f(3)+\dots +f(98)+f(99)&={\sqrt {1+1}}-{\sqrt {1}}+{\sqrt {2+1}}-{\sqrt {2}}+{\sqrt {3+1}}-{\sqrt {3}}+\cdot +{\sqrt {98+1}}-{\sqrt {98}}+{\sqrt {99+1}}-{\sqrt {99}}\\&={\sqrt {2}}-{\sqrt {1}}+{\sqrt {3}}-{\sqrt {2}}+{\sqrt {4}}-{\sqrt {3}}+\cdot +{\sqrt {99}}-{\sqrt {98}}+{\sqrt {100}}-{\sqrt {99}}\\&={\sqrt {100}}-{\sqrt {1}}\\&=10-1\\&=9\\\end{aligned}}}
Berapakah nilai dari
5
(
1
2025
+
2
2025
+
3
2025
+
⋯
+
2024
2025
)
{\displaystyle 5({\frac {1}{2025}}+{\frac {2}{2025}}+{\frac {3}{2025}}+\dots +{\frac {2024}{2025}})}
h
(
x
)
=
3
3
+
9
x
{\displaystyle h(x)={\frac {3}{3+9^{x}}}}
Jawaban
h
(
x
)
=
3
3
+
9
x
h
(
1
−
x
)
=
3
3
+
9
1
−
x
=
3
3
+
9
9
x
=
9
x
3
+
9
x
h
(
x
)
+
h
(
1
−
x
)
=
3
3
+
9
x
+
9
x
3
+
9
x
=
3
+
9
x
3
+
9
x
=
1
5
(
1
2025
+
2
2025
+
3
2025
+
⋯
+
(
1
−
2
2025
)
+
(
1
−
1
2025
)
)
5
(
1
+
1
+
1
+
⋯
+
1
+
1
)
sebanyak 1012 kali
5
(
1012
)
5060
{\displaystyle {\begin{aligned}h(x)&={\frac {3}{3+9^{x}}}\\h(1-x)&={\frac {3}{3+9^{1-x}}}\\&={\frac {3}{3+{\frac {9}{9^{x}}}}}\\&={\frac {9^{x}}{3+9^{x}}}\\h(x)+h(1-x)&={\frac {3}{3+9^{x}}}+{\frac {9^{x}}{3+9^{x}}}\\&={\frac {3+9^{x}}{3+9^{x}}}\\&=1\\&5({\frac {1}{2025}}+{\frac {2}{2025}}+{\frac {3}{2025}}+\dots +(1-{\frac {2}{2025}})+(1-{\frac {1}{2025}}))\\&5(1+1+1+\dots +1+1){\text{ sebanyak 1012 kali }}\\&5(1012)\\&5060\\\end{aligned}}}
Berapakah nilai dari
7
2025
−
7
2023
+
432
7
2024
+
7
2023
+
72
{\displaystyle {\frac {7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72}}}
Jawaban
7
2025
−
7
2023
+
432
7
2024
+
7
2023
+
72
=
7
2023
7
2
−
7
2023
+
48
×
9
7
2023
7
1
+
7
2023
+
8
×
9
=
7
2023
(
7
2
−
1
)
+
48
×
9
7
2023
(
7
1
+
1
)
+
8
×
9
=
7
2023
(
49
−
1
)
+
48
×
9
7
2023
(
7
+
1
)
+
8
×
9
=
7
2023
×
48
+
48
×
9
7
2023
×
8
+
8
×
9
=
48
(
7
2023
+
9
)
8
(
7
2023
+
9
)
=
48
8
=
6
{\displaystyle {\begin{aligned}{\frac {7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72}}&={\frac {7^{2023}7^{2}-7^{2023}+48\times 9}{7^{2023}7^{1}+7^{2023}+8\times 9}}\\&={\frac {7^{2023}(7^{2}-1)+48\times 9}{7^{2023}(7^{1}+1)+8\times 9}}\\&={\frac {7^{2023}(49-1)+48\times 9}{7^{2023}(7+1)+8\times 9}}\\&={\frac {7^{2023}\times 48+48\times 9}{7^{2023}\times 8+8\times 9}}\\&={\frac {48(7^{2023}+9)}{8(7^{2023}+9)}}\\&={\frac {48}{8}}\\&=6\\\end{aligned}}}
Berapakah nilai dari
t
a
n
(
x
+
π
4
)
{\displaystyle tan(x+{\frac {\pi }{4}})}
1
c
o
s
x
−
t
a
n
x
=
4
5
{\displaystyle {\frac {1}{cosx}}-tanx={\frac {4}{5}}}
Jawaban
1
c
o
s
x
−
t
a
n
x
=
4
5
s
e
c
x
−
t
a
n
x
=
4
5
s
e
c
2
x
−
t
a
n
2
x
=
1
(
s
e
c
x
+
t
a
n
x
)
(
s
e
c
x
−
t
a
n
x
)
=
1
(
s
e
c
x
+
t
a
n
x
)
4
5
=
1
s
e
c
x
+
t
a
n
x
=
5
4
kedua persamaan dengan cara metode eliminasi
2
t
a
n
x
=
5
4
−
4
5
2
t
a
n
x
=
9
20
t
a
n
x
=
9
40
t
a
n
(
x
+
π
4
)
=
t
a
n
x
+
t
a
n
π
4
1
−
t
a
n
x
⋅
t
a
n
π
4
=
9
40
+
1
1
−
9
40
⋅
1
=
49
40
31
40
=
49
31
{\displaystyle {\begin{aligned}{\frac {1}{cosx}}-tanx&={\frac {4}{5}}\\secx-tanx&={\frac {4}{5}}\\sec^{2}x-tan^{2}x&=1\\(secx+tanx)(secx-tanx)&=1\\(secx+tanx){\frac {4}{5}}&=1\\secx+tanx&={\frac {5}{4}}\\{\text{kedua persamaan dengan cara metode eliminasi }}\\2tanx&={\frac {5}{4}}-{\frac {4}{5}}\\2tanx&={\frac {9}{20}}\\tanx&={\frac {9}{40}}\\tan(x+{\frac {\pi }{4}})&={\frac {tanx+tan{\frac {\pi }{4}}}{1-tanx\cdot tan{\frac {\pi }{4}}}}\\&={\frac {{\frac {9}{40}}+1}{1-{\frac {9}{40}}\cdot 1}}\\&={\frac {\frac {49}{40}}{\frac {31}{40}}}\\&={\frac {49}{31}}\\\end{aligned}}}
Berapakah nilai dari
s
i
n
3
x
+
c
s
c
3
x
{\displaystyle sin^{3}x+csc^{3}x}
s
i
n
x
−
c
s
c
x
=
8
{\displaystyle sinx-cscx=8}
Jawaban
Dengan menggunakan rumus:
(
a
−
b
)
3
=
a
3
−
b
3
−
3
a
b
(
a
−
b
)
(
s
i
n
x
−
c
s
c
x
)
3
=
s
i
n
3
x
−
c
s
c
3
x
−
3
s
i
n
x
c
s
c
x
(
s
i
n
x
−
c
s
c
x
)
8
3
=
s
i
n
3
x
−
c
s
c
3
x
−
3
s
i
n
x
(
1
s
i
n
x
)
(
8
)
512
=
s
i
n
3
x
−
c
s
c
3
x
−
24
s
i
n
3
x
−
c
s
c
3
x
=
512
+
24
s
i
n
3
x
−
c
s
c
3
x
=
536
{\displaystyle {\begin{aligned}{\text{ Dengan menggunakan rumus: }}(a-b)^{3}&=a^{3}-b^{3}-3ab(a-b)\\(sinx-cscx)^{3}&=sin^{3}x-csc^{3}x-3sinxcscx(sinx-cscx)\\8^{3}&=sin^{3}x-csc^{3}x-3sinx({\frac {1}{sinx}})(8)\\512&=sin^{3}x-csc^{3}x-24\\sin^{3}x-csc^{3}x&=512+24\\sin^{3}x-csc^{3}x&=536\\\end{aligned}}}
Berapakah nilai dari
(
s
i
n
x
+
1
c
o
s
x
)
2
+
(
c
o
s
x
+
1
s
i
n
x
)
2
{\displaystyle (sinx+{\frac {1}{cosx}})^{2}+(cosx+{\frac {1}{sinx}})^{2}}
1
s
i
n
x
+
1
c
o
s
x
=
10
{\displaystyle {\frac {1}{sinx}}+{\frac {1}{cosx}}=10}
Jawaban
1
s
i
n
x
+
1
c
o
s
x
=
10
1
s
i
n
2
x
+
2
s
i
n
x
⋅
c
o
s
x
+
1
c
o
s
2
x
=
100
(
s
i
n
x
+
1
c
o
s
x
)
2
+
(
c
o
s
x
+
1
s
i
n
x
)
2
=
s
i
n
2
x
+
2
s
i
n
x
c
o
s
x
+
1
c
o
s
2
x
+
c
o
s
2
x
+
2
c
o
s
x
s
i
n
x
+
1
s
i
n
2
x
=
1
+
1
s
i
n
2
x
+
2
(
s
i
n
2
x
+
c
o
s
2
x
)
s
i
n
x
⋅
c
o
s
x
+
1
c
o
s
2
x
=
1
+
1
s
i
n
2
x
+
2
s
i
n
x
⋅
c
o
s
x
+
1
c
o
s
2
x
=
1
+
100
=
101
{\displaystyle {\begin{aligned}{\frac {1}{sinx}}+{\frac {1}{cosx}}&=10\\{\frac {1}{sin^{2}x}}+{\frac {2}{sinx\cdot cosx}}+{\frac {1}{cos^{2}x}}&=100\\(sinx+{\frac {1}{cosx}})^{2}+(cosx+{\frac {1}{sinx}})^{2}&=sin^{2}x+{\frac {2sinx}{cosx}}+{\frac {1}{cos^{2}x}}+cos^{2}x+{\frac {2cosx}{sinx}}+{\frac {1}{sin^{2}x}}\\&=1+{\frac {1}{sin^{2}x}}+{\frac {2(sin^{2}x+cos^{2}x)}{sinx\cdot cosx}}+{\frac {1}{cos^{2}x}}\\&=1+{\frac {1}{sin^{2}x}}+{\frac {2}{sinx\cdot cosx}}+{\frac {1}{cos^{2}x}}\\&=1+100\\&=101\\\end{aligned}}}
Berapakah nilai dari (x-1)6 jika
x
=
4
c
o
s
55
∘
c
o
s
25
∘
c
o
s
10
∘
+
s
i
n
40
∘
s
i
n
80
∘
{\displaystyle x={\frac {4cos55^{\circ }cos25^{\circ }cos10^{\circ }+sin40^{\circ }}{sin80^{\circ }}}}
Jawaban
s
i
n
80
∘
=
c
o
s
10
∘
s
i
n
80
∘
−
c
o
s
10
∘
=
0
x
=
4
c
o
s
55
∘
c
o
s
25
∘
c
o
s
10
∘
+
s
i
n
40
∘
s
i
n
80
∘
=
4
c
o
s
55
∘
c
o
s
25
∘
c
o
s
10
∘
+
2
s
i
n
20
∘
c
o
s
20
∘
c
o
s
10
∘
=
4
c
o
s
55
∘
c
o
s
25
∘
c
o
s
10
∘
+
4
s
i
n
10
∘
c
o
s
10
∘
c
o
s
20
∘
c
o
s
10
∘
=
4
c
o
s
55
∘
c
o
s
25
∘
+
4
s
i
n
10
∘
c
o
s
20
∘
=
2
(
2
c
o
s
55
∘
c
o
s
25
∘
+
2
s
i
n
10
∘
c
o
s
20
∘
)
=
2
(
c
o
s
80
∘
+
c
o
s
30
∘
+
s
i
n
30
∘
+
s
i
n
(
−
10
)
∘
)
=
2
(
c
o
s
80
∘
+
c
o
s
30
∘
+
s
i
n
30
∘
−
s
i
n
10
∘
)
=
2
(
c
o
s
80
∘
−
s
i
n
10
∘
+
c
o
s
30
∘
+
s
i
n
30
∘
)
=
2
(
c
o
s
80
∘
−
s
i
n
(
90
∘
−
80
∘
)
+
3
2
+
1
2
)
=
2
(
c
o
s
80
∘
−
c
o
s
80
∘
+
3
2
+
1
2
)
=
2
(
3
2
+
1
2
)
=
3
+
1
x
−
1
=
3
(
x
−
1
)
6
=
(
3
)
6
=
27
{\displaystyle {\begin{aligned}sin80^{\circ }&=cos10^{\circ }\\sin80^{\circ }-cos10^{\circ }&=0\\x&={\frac {4cos55^{\circ }cos25^{\circ }cos10^{\circ }+sin40^{\circ }}{sin80^{\circ }}}\\&={\frac {4cos55^{\circ }cos25^{\circ }cos10^{\circ }+2sin20^{\circ }cos20^{\circ }}{cos10^{\circ }}}\\&={\frac {4cos55^{\circ }cos25^{\circ }cos10^{\circ }+4sin10^{\circ }cos10^{\circ }cos20^{\circ }}{cos10^{\circ }}}\\&=4cos55^{\circ }cos25^{\circ }+4sin10^{\circ }cos20^{\circ }\\&=2(2cos55^{\circ }cos25^{\circ }+2sin10^{\circ }cos20^{\circ })\\&=2(cos80^{\circ }+cos30^{\circ }+sin30^{\circ }+sin(-10)^{\circ })\\&=2(cos80^{\circ }+cos30^{\circ }+sin30^{\circ }-sin10^{\circ })\\&=2(cos80^{\circ }-sin10^{\circ }+cos30^{\circ }+sin30^{\circ })\\&=2(cos80^{\circ }-sin(90^{\circ }-80^{\circ })+{\frac {\sqrt {3}}{2}}+{\frac {1}{2}})\\&=2(cos80^{\circ }-cos80^{\circ }+{\frac {\sqrt {3}}{2}}+{\frac {1}{2}})\\&=2({\frac {\sqrt {3}}{2}}+{\frac {1}{2}})\\&={\sqrt {3}}+1\\x-1&={\sqrt {3}}\\(x-1)^{6}&=({\sqrt {3}})^{6}\\&=27\\\end{aligned}}}
Berapakah nilai dari x jika
x
=
x
s
i
n
20
∘
−
x
2
s
i
n
10
∘
2
s
i
n
20
∘
−
s
i
n
40
∘
{\displaystyle x={\frac {xsin20^{\circ }-x^{2}sin10^{\circ }}{2sin20^{\circ }-sin40^{\circ }}}}
Jawaban
x
=
x
s
i
n
20
∘
−
x
2
s
i
n
10
∘
2
s
i
n
20
∘
−
s
i
n
40
∘
2
x
s
i
n
20
∘
−
x
s
i
n
40
∘
=
x
s
i
n
20
∘
−
x
2
s
i
n
10
∘
x
2
s
i
n
10
∘
+
x
s
i
n
20
∘
−
x
s
i
n
40
∘
=
0
x
(
x
s
i
n
10
∘
+
s
i
n
20
∘
−
s
i
n
40
∘
)
=
0
x
=
0
atau
x
s
i
n
10
∘
+
s
i
n
20
∘
−
s
i
n
40
∘
=
0
x
s
i
n
10
∘
+
s
i
n
20
∘
−
s
i
n
40
∘
=
0
x
s
i
n
10
∘
=
s
i
n
40
∘
−
s
i
n
20
∘
x
=
s
i
n
40
∘
−
s
i
n
20
∘
s
i
n
10
∘
=
2
c
o
s
30
∘
s
i
n
10
∘
s
i
n
10
∘
=
2
c
o
s
30
∘
=
2
3
2
=
3
{\displaystyle {\begin{aligned}x&={\frac {xsin20^{\circ }-x^{2}sin10^{\circ }}{2sin20^{\circ }-sin40^{\circ }}}\\2xsin20^{\circ }-xsin40^{\circ }&=xsin20^{\circ }-x^{2}sin10^{\circ }\\x^{2}sin10^{\circ }+xsin20^{\circ }-xsin40^{\circ }&=0\\x(xsin10^{\circ }+sin20^{\circ }-sin40^{\circ })&=0\\x=0&{\text{ atau }}xsin10^{\circ }+sin20^{\circ }-sin40^{\circ }=0\\xsin10^{\circ }+sin20^{\circ }-sin40^{\circ }&=0\\xsin10^{\circ }&=sin40^{\circ }-sin20^{\circ }\\x&={\frac {sin40^{\circ }-sin20^{\circ }}{sin10^{\circ }}}\\&={\frac {2cos30^{\circ }sin10^{\circ }}{sin10^{\circ }}}\\&=2cos30^{\circ }\\&={\frac {2{\sqrt {3}}}{2}}\\&={\sqrt {3}}\\\end{aligned}}}
Berapakah nilai dari
x
y
{\displaystyle {\frac {x}{y}}}
x
2
x
2
−
16
y
2
=
625
49
{\displaystyle {\frac {x^{2}}{x^{2}-16y^{2}}}={\frac {625}{49}}}
Jawaban
x
2
x
2
−
16
y
2
=
625
49
x
2
−
16
y
2
x
2
=
49
625
(terbalik posisinya)
1
−
16
y
2
x
2
=
49
625
16
y
2
x
2
=
1
−
49
625
(
4
y
x
)
2
=
576
625
(
4
y
x
)
2
=
(
24
25
)
2
4
y
x
=
24
25
y
x
=
6
25
x
y
=
25
6
{\displaystyle {\begin{aligned}{\frac {x^{2}}{x^{2}-16y^{2}}}&={\frac {625}{49}}\\{\frac {x^{2}-16y^{2}}{x^{2}}}&={\frac {49}{625}}{\text{ (terbalik posisinya)}}\\1-{\frac {16y^{2}}{x^{2}}}&={\frac {49}{625}}\\{\frac {16y^{2}}{x^{2}}}&=1-{\frac {49}{625}}\\({\frac {4y}{x}})^{2}&={\frac {576}{625}}\\({\frac {4y}{x}})^{2}&=({\frac {24}{25}})^{2}\\{\frac {4y}{x}}&={\frac {24}{25}}\\{\frac {y}{x}}&={\frac {6}{25}}\\{\frac {x}{y}}&={\frac {25}{6}}\\\end{aligned}}}
Berapakah nilai dari xy jika
x
4
+
y
4
+
x
2
y
2
=
15
dan
x
2
+
y
2
+
x
y
=
5
{\displaystyle x^{4}+y^{4}+x^{2}y^{2}=15{\text{ dan }}x^{2}+y^{2}+xy=5}
Jawaban
x
2
+
y
2
+
x
y
=
5
x
2
+
y
2
=
5
−
x
y
x
4
+
y
4
+
x
2
y
2
=
15
(
x
2
)
2
+
(
y
2
)
2
+
2
x
2
y
2
−
x
2
y
2
=
15
(
x
2
+
y
2
)
2
−
x
2
y
2
=
15
(
5
−
x
y
)
2
−
x
2
y
2
=
15
25
−
10
x
y
+
x
2
y
2
−
x
2
y
2
=
15
25
−
10
x
y
=
15
10
x
y
=
10
x
y
=
1
{\displaystyle {\begin{aligned}x^{2}+y^{2}+xy&=5\\x^{2}+y^{2}&=5-xy\\x^{4}+y^{4}+x^{2}y^{2}&=15\\(x^{2})^{2}+(y^{2})^{2}+2x^{2}y^{2}-x^{2}y^{2}&=15\\(x^{2}+y^{2})^{2}-x^{2}y^{2}&=15\\(5-xy)^{2}-x^{2}y^{2}&=15\\25-10xy+x^{2}y^{2}-x^{2}y^{2}&=15\\25-10xy&=15\\10xy&=10\\xy&=1\\\end{aligned}}}
Berapakah nilai dari x jika
4
x
=
63
(
4
3
+
1
)
(
4
6
+
1
)
(
4
12
+
1
)
+
1
{\displaystyle 4^{x}=63(4^{3}+1)(4^{6}+1)(4^{12}+1)+1}
Jawaban
4
x
=
63
(
4
3
+
1
)
(
4
6
+
1
)
(
4
12
+
1
)
+
1
4
x
−
1
=
63
(
4
3
+
1
)
(
4
6
+
1
)
(
4
12
+
1
)
=
63
(
4
3
+
1
)
(
4
6
+
1
)
(
4
12
+
1
)
4
3
−
1
4
3
−
1
=
63
(
4
3
+
1
)
(
4
6
+
1
)
(
4
12
+
1
)
4
3
−
1
63
=
(
4
3
+
1
)
(
4
6
+
1
)
(
4
12
+
1
)
(
4
3
−
1
)
=
(
4
3
−
1
)
(
4
3
+
1
)
(
4
6
+
1
)
(
4
12
+
1
)
=
(
4
6
−
1
)
(
4
6
+
1
)
(
4
12
+
1
)
=
(
4
12
−
1
)
(
4
12
+
1
)
=
4
24
−
1
4
x
=
4
24
x
=
24
{\displaystyle {\begin{aligned}4^{x}&=63(4^{3}+1)(4^{6}+1)(4^{12}+1)+1\\4^{x}-1&=63(4^{3}+1)(4^{6}+1)(4^{12}+1)\\&=63(4^{3}+1)(4^{6}+1)(4^{12}+1){\frac {4^{3}-1}{4^{3}-1}}\\&=63(4^{3}+1)(4^{6}+1)(4^{12}+1){\frac {4^{3}-1}{63}}\\&=(4^{3}+1)(4^{6}+1)(4^{12}+1)(4^{3}-1)\\&=(4^{3}-1)(4^{3}+1)(4^{6}+1)(4^{12}+1)\\&=(4^{6}-1)(4^{6}+1)(4^{12}+1)\\&=(4^{12}-1)(4^{12}+1)\\&=4^{24}-1\\4^{x}&=4^{24}\\x&=24\\\end{aligned}}}
Berapakah nilai dari
x
4
−
5
x
3
+
2
x
2
+
5
x
+
3
x
2
−
4
x
+
1
{\displaystyle {\frac {x^{4}-5x^{3}+2x^{2}+5x+3}{x^{2}-4x+1}}}
x
=
9
+
4
5
{\displaystyle x={\sqrt {9+4{\sqrt {5}}}}}
Jawaban
x
=
9
+
4
5
x
=
2
+
5
x
2
=
9
+
4
5
x
2
−
4
x
=
9
+
4
5
−
4
(
2
+
5
)
x
2
−
4
x
=
1
x
2
=
4
x
+
1
x
3
=
x
⋅
x
2
=
x
(
4
x
+
1
)
=
4
x
2
+
x
=
4
(
4
x
+
1
)
+
x
=
16
x
+
4
+
x
=
17
x
+
4
x
4
=
x
⋅
x
3
=
x
(
17
x
+
4
)
=
17
x
2
+
4
x
=
17
(
4
x
+
1
)
+
4
x
=
68
x
+
17
+
4
x
=
72
x
+
17
x
4
−
5
x
3
+
2
x
2
+
5
x
+
3
x
2
−
4
x
+
1
=
72
x
+
17
−
5
(
17
x
+
4
)
+
2
(
4
x
+
1
)
+
5
x
+
3
1
+
1
=
72
x
+
17
−
85
x
−
20
+
8
x
+
2
+
5
x
+
3
2
=
2
2
=
1
{\displaystyle {\begin{aligned}x&={\sqrt {9+4{\sqrt {5}}}}\\x&=2+{\sqrt {5}}\\x^{2}&=9+4{\sqrt {5}}\\x^{2}-4x&=9+4{\sqrt {5}}-4(2+{\sqrt {5}})\\x^{2}-4x&=1\\x^{2}&=4x+1\\x^{3}&=x\cdot x^{2}\\&=x(4x+1)\\&=4x^{2}+x\\&=4(4x+1)+x\\&=16x+4+x\\&=17x+4\\x^{4}&=x\cdot x^{3}\\&=x(17x+4)\\&=17x^{2}+4x\\&=17(4x+1)+4x\\&=68x+17+4x\\&=72x+17\\{\frac {x^{4}-5x^{3}+2x^{2}+5x+3}{x^{2}-4x+1}}&={\frac {72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1}}\\&={\frac {72x+17-85x-20+8x+2+5x+3}{2}}\\&={\frac {2}{2}}\\&=1\\\end{aligned}}}
Berapakah nilai dari
x
3
+
1
x
5
−
x
4
−
x
3
+
x
2
{\displaystyle {\sqrt {\frac {x^{3}+1}{x^{5}-x^{4}-x^{3}+x^{2}}}}}
61
{\displaystyle {\sqrt {61}}}
Jawaban
misalkan
x
3
+
1
x
5
−
x
4
−
x
3
+
x
2
=
p
p
=
x
3
+
1
x
5
−
x
4
−
x
3
+
x
2
=
x
3
+
1
x
5
−
x
4
−
(
x
3
−
x
2
)
=
x
3
+
1
x
4
(
x
−
1
)
−
x
2
(
x
−
1
)
=
(
x
+
1
)
(
x
2
−
x
+
1
)
x
4
(
x
−
1
)
−
x
2
(
x
−
1
)
=
(
x
+
1
)
(
x
2
−
x
+
1
)
(
x
−
1
)
(
x
4
−
x
2
)
=
(
x
+
1
)
(
x
2
−
x
+
1
)
(
x
−
1
)
x
2
(
x
2
−
1
)
=
(
x
+
1
)
(
x
2
−
x
+
1
)
(
x
−
1
)
x
2
(
x
−
1
)
(
x
+
1
)
=
x
2
−
x
+
1
x
2
(
x
−
1
)
2
=
x
2
−
x
+
1
(
x
(
x
−
1
)
)
2
=
x
(
x
−
1
)
+
1
(
x
(
x
−
1
)
)
2
2
x
−
1
=
61
x
=
61
+
1
2
x
−
1
=
61
−
1
2
x
(
x
−
1
)
=
(
61
+
1
2
)
(
61
−
1
2
)
=
61
−
1
4
=
60
4
=
15
p
=
x
(
x
−
1
)
+
1
(
x
(
x
−
1
)
)
2
=
15
+
1
15
2
=
16
15
2
p
=
16
15
2
=
4
15
{\displaystyle {\begin{aligned}{\text{misalkan }}{\frac {x^{3}+1}{x^{5}-x^{4}-x^{3}+x^{2}}}=p\\p&={\frac {x^{3}+1}{x^{5}-x^{4}-x^{3}+x^{2}}}\\&={\frac {x^{3}+1}{x^{5}-x^{4}-(x^{3}-x^{2})}}\\&={\frac {x^{3}+1}{x^{4}(x-1)-x^{2}(x-1)}}\\&={\frac {(x+1)(x^{2}-x+1)}{x^{4}(x-1)-x^{2}(x-1)}}\\&={\frac {(x+1)(x^{2}-x+1)}{(x-1)(x^{4}-x^{2})}}\\&={\frac {(x+1)(x^{2}-x+1)}{(x-1)x^{2}(x^{2}-1)}}\\&={\frac {(x+1)(x^{2}-x+1)}{(x-1)x^{2}(x-1)(x+1)}}\\&={\frac {x^{2}-x+1}{x^{2}(x-1)^{2}}}\\&={\frac {x^{2}-x+1}{(x(x-1))^{2}}}\\&={\frac {x(x-1)+1}{(x(x-1))^{2}}}\\2x-1&={\sqrt {61}}\\x&={\frac {{\sqrt {61}}+1}{2}}\\x-1&={\frac {{\sqrt {61}}-1}{2}}\\x(x-1)&=({\frac {{\sqrt {61}}+1}{2}})({\frac {{\sqrt {61}}-1}{2}})\\&={\frac {61-1}{4}}\\&={\frac {60}{4}}\\&=15\\p&={\frac {x(x-1)+1}{(x(x-1))^{2}}}\\&={\frac {15+1}{15^{2}}}\\&={\frac {16}{15^{2}}}\\{\sqrt {p}}&={\sqrt {\frac {16}{15^{2}}}}\\&={\frac {4}{15}}\\\end{aligned}}}
Berapakah nilai dari
(
x
−
3
x
)
25
{\displaystyle ({\frac {x-3}{x}})^{25}}
x
+
8
5
+
2
5
=
1
+
16
5
+
4
5
{\displaystyle x+{\sqrt[{5}]{8}}+{\sqrt[{5}]{2}}=1+{\sqrt[{5}]{16}}+{\sqrt[{5}]{4}}}
Jawaban
x
+
8
5
+
2
5
=
1
+
16
5
+
4
5
x
+
(
2
5
)
3
+
2
5
=
1
+
(
2
5
)
4
+
(
2
5
)
2
x
=
(
2
5
)
4
−
(
2
5
)
3
+
(
2
5
)
2
−
2
5
+
1
misalkan
2
5
=
p
x
=
p
4
−
p
3
+
p
2
−
p
+
1
x
=
p
5
+
1
p
+
1
(
x
−
3
x
)
25
=
(
1
−
3
x
)
25
=
(
1
−
3
p
5
+
1
p
+
1
)
25
=
(
1
−
3
(
p
+
1
)
p
5
+
1
)
25
=
(
1
−
3
(
2
5
+
1
)
(
2
5
)
5
+
1
)
25
=
(
1
−
(
3
2
5
+
3
)
2
+
1
)
25
=
(
1
−
(
3
2
5
+
3
)
3
)
25
=
(
3
−
(
3
2
5
+
3
)
3
)
25
=
(
3
−
3
2
5
−
3
)
3
)
25
=
(
−
2
5
)
25
=
(
−
2
)
5
=
−
32
{\displaystyle {\begin{aligned}x+{\sqrt[{5}]{8}}+{\sqrt[{5}]{2}}&=1+{\sqrt[{5}]{16}}+{\sqrt[{5}]{4}}\\x+({\sqrt[{5}]{2}})^{3}+{\sqrt[{5}]{2}}&=1+({\sqrt[{5}]{2}})^{4}+({\sqrt[{5}]{2}})^{2}\\x&=({\sqrt[{5}]{2}})^{4}-({\sqrt[{5}]{2}})^{3}+({\sqrt[{5}]{2}})^{2}-{\sqrt[{5}]{2}}+1\\{\text{misalkan }}{\sqrt[{5}]{2}}=p\\x&=p^{4}-p^{3}+p^{2}-p+1\\x&={\frac {p^{5}+1}{p+1}}\\({\frac {x-3}{x}})^{25}&=(1-{\frac {3}{x}})^{25}\\&=(1-{\frac {3}{\frac {p^{5}+1}{p+1}}})^{25}\\&=(1-{\frac {3(p+1)}{p^{5}+1}})^{25}\\&=(1-{\frac {3({\sqrt[{5}]{2}}+1)}{({\sqrt[{5}]{2}})^{5}+1}})^{25}\\&=(1-{\frac {(3{\sqrt[{5}]{2}}+3)}{2+1}})^{25}\\&=(1-{\frac {(3{\sqrt[{5}]{2}}+3)}{3}})^{25}\\&=({\frac {3-(3{\sqrt[{5}]{2}}+3)}{3}})^{25}\\&=({\frac {3-3{\sqrt[{5}]{2}}-3)}{3}})^{25}\\&=(-{\sqrt[{5}]{2}})^{25}\\&=(-2)^{5}\\&=-32\\\end{aligned}}}
Berapakah nilai dari
x
50
+
x
49
+
x
48
+
x
47
+
x
46
{\displaystyle x^{50}+x^{49}+x^{48}+x^{47}+x^{46}}
x
2
+
x
+
1
=
0
{\displaystyle x^{2}+x+1=0}
Jawaban
x
2
+
x
+
1
=
0
x
2
+
x
=
−
1
x
3
−
1
x
−
1
=
0
x
3
=
1
x
=
1
x
50
+
x
49
+
x
48
+
x
47
+
x
46
=
x
48
(
x
2
+
x
+
1
)
+
x
45
(
x
2
+
x
)
=
x
48
(
0
)
+
(
x
3
)
15
(
−
1
)
=
0
+
(
1
)
15
(
−
1
)
=
−
1
{\displaystyle {\begin{aligned}x^{2}+x+1&=0\\x^{2}+x&=-1\\{\frac {x^{3}-1}{x-1}}&=0\\x^{3}&=1\\x&=1\\x^{50}+x^{49}+x^{48}+x^{47}+x^{46}&=x^{48}(x^{2}+x+1)+x^{45}(x^{2}+x)\\&=x^{48}(0)+(x^{3})^{15}(-1)\\&=0+(1)^{15}(-1)\\&=-1\\\end{aligned}}}
Berapakah nilai dari
x
42
+
x
36
+
x
30
+
x
24
+
x
18
+
x
12
+
x
6
+
1
{\displaystyle x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1}
x
+
1
x
=
3
{\displaystyle x+{\frac {1}{x}}={\sqrt {3}}}
Jawaban
x
+
1
x
=
3
x
2
+
2
+
1
x
2
=
3
x
2
−
1
+
1
x
2
=
0
x
2
(
x
2
−
1
+
1
x
2
)
=
x
2
(
0
)
x
4
−
x
2
+
1
=
0
(
x
2
+
1
)
(
x
4
−
x
2
+
1
)
=
(
x
2
+
1
)
0
x
6
−
x
4
+
x
2
+
x
4
−
x
2
+
1
=
0
x
6
+
1
=
0
x
6
=
−
1
x
42
+
x
36
+
x
30
+
x
24
+
x
18
+
x
12
+
x
6
+
1
=
x
6
7
+
x
6
6
+
x
6
5
+
x
6
4
+
x
6
3
+
x
6
2
+
x
6
+
1
=
(
−
1
)
7
+
(
−
1
)
6
+
(
−
1
)
5
+
(
−
1
)
4
+
(
−
1
)
3
+
(
−
1
)
2
−
1
+
1
=
−
1
+
1
−
1
+
1
−
1
+
1
−
1
+
1
=
0
{\displaystyle {\begin{aligned}x+{\frac {1}{x}}&={\sqrt {3}}\\x^{2}+2+{\frac {1}{x^{2}}}&=3\\x^{2}-1+{\frac {1}{x^{2}}}&=0\\x^{2}(x^{2}-1+{\frac {1}{x^{2}}})&=x^{2}(0)\\x^{4}-x^{2}+1&=0\\(x^{2}+1)(x^{4}-x^{2}+1)&=(x^{2}+1)0\\x^{6}-x^{4}+x^{2}+x^{4}-x^{2}+1&=0\\x^{6}+1&=0\\x^{6}&=-1\\x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1&={x^{6}}^{7}+{x^{6}}^{6}+{x^{6}}^{5}+{x^{6}}^{4}+{x^{6}}^{3}+{x^{6}}^{2}+x^{6}+1\\&=(-1)^{7}+(-1)^{6}+(-1)^{5}+(-1)^{4}+(-1)^{3}+(-1)^{2}-1+1\\&=-1+1-1+1-1+1-1+1\\&=0\\\end{aligned}}}
Diberikan fungsi kuadrat f(x)=ax2 +bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari
c
−
b
a
−
1
{\displaystyle {\frac {c-b}{a-1}}}
Jawaban
f
(
x
)
=
a
x
2
+
b
x
+
c
f
(
2
)
=
a
(
2
)
2
+
2
b
+
c
=
4
=
4
a
+
2
b
+
c
=
4
f
(
7
)
=
a
(
7
)
2
+
7
b
+
c
=
49
=
49
a
+
7
b
+
c
=
49
49
a
+
7
b
+
c
=
49
4
a
+
2
b
+
c
=
4
45
a
+
5
b
=
45
(f(7) dikurangi f(2))
9
a
+
b
=
9
b
=
−
9
a
+
9
4
a
+
2
b
+
c
=
4
4
a
+
2
(
−
9
a
+
9
)
+
c
=
4
4
a
−
18
a
+
18
+
c
=
4
−
14
a
+
18
+
c
=
4
c
=
14
a
−
14
c
−
b
a
−
1
=
14
a
−
14
−
(
−
9
a
+
9
)
a
−
1
=
14
(
a
−
1
)
+
9
(
a
−
1
)
a
−
1
=
(
14
+
9
)
(
a
−
1
)
a
−
1
=
23
{\displaystyle {\begin{aligned}f(x)&=ax^{2}+bx+c\\f(2)&=a(2)^{2}+2b+c=4\\&=4a+2b+c=4\\f(7)&=a(7)^{2}+7b+c=49\\&=49a+7b+c=49\\49a+7b+c&=49\\4a+2b+c&=4\\45a+5b&=45{\text{ (f(7) dikurangi f(2)) }}\\9a+b&=9\\b&=-9a+9\\4a+2b+c&=4\\4a+2(-9a+9)+c&=4\\4a-18a+18+c&=4\\-14a+18+c&=4\\c&=14a-14\\{\frac {c-b}{a-1}}&={\frac {14a-14-(-9a+9)}{a-1}}\\&={\frac {14(a-1)+9(a-1)}{a-1}}\\&={\frac {(14+9)(a-1)}{a-1}}\\&=23\\\end{aligned}}}
Jika a3 +b3 = 242 dan a+b = 11 maka berapa hasil dari (a-b)2 ? Jawaban
(
a
+
b
)
3
=
a
3
+
b
3
+
3
a
b
(
a
+
b
)
11
3
=
242
+
3
a
b
(
11
)
(dibagi 11)
11
2
=
22
+
3
a
b
121
=
22
+
3
a
b
99
=
3
a
b
a
b
=
33
(
a
−
b
)
2
=
a
2
+
b
2
−
2
a
b
=
(
(
a
+
b
)
2
−
2
a
b
)
−
2
a
b
=
(
a
+
b
)
2
−
4
a
b
=
11
2
−
4
(
33
)
=
121
−
132
=
−
11
{\displaystyle {\begin{aligned}(a+b)^{3}&=a^{3}+b^{3}+3ab(a+b)\\11^{3}&=242+3ab(11){\text{ (dibagi 11)}}\\11^{2}&=22+3ab\\121&=22+3ab\\99&=3ab\\ab&=33\\(a-b)^{2}&=a^{2}+b^{2}-2ab\\&=((a+b)^{2}-2ab)-2ab\\&=(a+b)^{2}-4ab\\&=11^{2}-4(33)\\&=121-132\\&=-11\\\end{aligned}}}
Misalkan f(x) adalah fungsi yang berlaku ∀x ∈ R sebagai berikut: f(x)+f(15-x) = 2024
f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)!
Jawaban
f
(
x
)
+
f
(
15
−
x
)
=
2024
f
(
15
+
x
)
=
f
(
x
)
+
2020
∗
cara 1
ganti x dengan 15+x
f
(
15
+
x
)
+
f
(
−
x
)
=
2024
f
(
15
+
x
)
−
f
(
x
)
=
2020
persamaan 1 dan 2 dihasilkan sebagai berikut
f
(
x
)
+
f
(
−
x
)
=
4
lalu dikalikan 2 masing-masing menjadi
2
f
(
x
)
+
2
f
(
−
x
)
=
8
maka
2
f
(
2025
)
+
2
f
(
−
2025
)
=
8
∗
cara 2
ganti x dengan -x
f
(
−
x
)
+
f
(
15
+
x
)
=
2024
f
(
15
+
x
)
−
f
(
x
)
=
2020
persamaan 1 dan 2 dihasilkan sebagai berikut
f
(
x
)
+
f
(
−
x
)
=
4
lalu dikalikan 2 masing-masing menjadi
2
f
(
x
)
+
2
f
(
−
x
)
=
8
maka
2
f
(
2025
)
+
2
f
(
−
2025
)
=
8
{\displaystyle {\begin{aligned}f(x)+f(15-x)&=2024\\f(15+x)&=f(x)+2020\\*{\text{cara 1 }}\\{\text{ganti x dengan 15+x }}\\f(15+x)+f(-x)&=2024\\f(15+x)-f(x)&=2020\\{\text{persamaan 1 dan 2 dihasilkan sebagai berikut }}\\f(x)+f(-x)&=4\\{\text{lalu dikalikan 2 masing-masing menjadi }}\\2f(x)+2f(-x)=8\\{\text{maka }}2f(2025)+2f(-2025)=8\\*{\text{cara 2 }}\\{\text{ganti x dengan -x }}\\f(-x)+f(15+x)&=2024\\f(15+x)-f(x)&=2020\\{\text{persamaan 1 dan 2 dihasilkan sebagai berikut }}\\f(x)+f(-x)&=4\\{\text{lalu dikalikan 2 masing-masing menjadi }}\\2f(x)+2f(-x)=8\\{\text{maka }}2f(2025)+2f(-2025)=8\\\end{aligned}}}
Misalkan f suatu fungsi yang memenuhi
f
(
1
x
)
+
1
x
f
(
−
x
)
=
3
x
{\displaystyle f({\frac {1}{x}})+{\frac {1}{x}}f(-x)=3x}
Jawaban
f
(
1
x
)
+
1
x
f
(
−
x
)
=
3
x
ganti x dengan 1/3
f
(
3
)
+
3
f
(
−
1
3
)
=
1
ganti x dengan -3
f
(
−
1
3
)
−
1
3
f
(
3
)
=
−
9
dikalikan 3
3
f
(
−
1
3
)
−
f
(
3
)
=
−
27
persamaan 1 dan 2 dihasilkan sebagai berikut
2
f
(
3
)
=
28
f
(
3
)
=
14
{\displaystyle {\begin{aligned}f({\frac {1}{x}})+{\frac {1}{x}}f(-x)&=3x\\{\text{ganti x dengan 1/3 }}\\f(3)+3f(-{\frac {1}{3}})&=1\\{\text{ganti x dengan -3 }}\\f(-{\frac {1}{3}})-{\frac {1}{3}}f(3)=-9\\{\text{dikalikan 3 }}\\3f(-{\frac {1}{3}})-f(3)&=-27\\{\text{persamaan 1 dan 2 dihasilkan sebagai berikut }}\\2f(3)&=28\\f(3)&=14\\\end{aligned}}}
Diketahui polinom
f
(
7
b
−
1
)
=
7
3
b
−
10
{\displaystyle f(7^{b}-1)=7^{3b}-10}
Jawaban
∗
c
a
r
a
1
f
(
5
)
=
f
(
7
b
−
1
)
5
=
7
b
−
1
7
b
=
6
f
(
7
b
−
1
)
=
7
3
b
−
10
=
(
7
b
)
3
−
10
f
(
6
−
1
)
=
6
3
−
10
f
(
5
)
=
216
−
10
=
206
∗
c
a
r
a
2
misalkan
7
b
−
1
=
a
maka
7
b
=
a
+
1
f
(
7
b
−
1
)
=
7
3
b
−
10
=
(
7
b
)
3
−
10
f
(
a
)
=
(
a
+
1
)
3
−
10
f
(
5
)
=
(
5
+
1
)
3
−
10
=
6
3
−
10
=
216
−
10
=
206
{\displaystyle {\begin{aligned}*cara1\\f(5)&=f(7^{b}-1)\\5&=7^{b}-1\\7^{b}&=6\\f(7^{b}-1)&=7^{3b}-10\\&=(7^{b})^{3}-10\\f(6-1)&=6^{3}-10\\f(5)&=216-10\\&=206\\*cara2\\{\text{misalkan }}7^{b}-1=a{\text{ maka }}7^{b}=a+1\\f(7^{b}-1)&=7^{3b}-10\\&=(7^{b})^{3}-10\\f(a)&=(a+1)^{3}-10\\f(5)&=(5+1)^{3}-10\\&=6^{3}-10\\&=216-10\\&=206\\\end{aligned}}}
Diketahui polinom
f
(
6
b
−
7
)
=
6
3
b
−
2
⋅
6
2
b
−
4
{\displaystyle f(6^{b}-7)=6^{3b}-2\cdot 6^{2b}-4}
Jawaban
∗
c
a
r
a
1
f
(
−
2
)
=
f
(
6
b
−
7
)
−
2
=
6
b
−
7
6
b
=
5
f
(
6
b
−
7
)
=
6
3
b
−
2
⋅
6
2
b
−
4
=
(
6
b
)
3
−
2
⋅
(
6
b
)
2
−
4
f
(
5
−
7
)
=
5
3
−
2
⋅
5
2
−
4
f
(
−
2
)
=
125
−
50
−
4
=
71
∗
c
a
r
a
2
misalkan
6
b
−
7
=
a
maka
6
b
=
a
+
7
f
(
6
b
−
7
)
=
6
3
b
−
2
⋅
6
2
b
−
4
=
(
6
b
)
3
−
2
⋅
(
6
b
)
2
−
4
f
(
a
)
=
(
a
+
7
)
3
−
2
(
a
+
7
)
2
−
4
f
(
−
2
)
=
(
−
2
+
7
)
3
−
2
(
−
2
+
7
)
2
−
4
=
5
3
−
2
(
5
)
2
−
4
=
125
−
50
−
4
=
71
{\displaystyle {\begin{aligned}*cara1\\f(-2)&=f(6^{b}-7)\\-2&=6^{b}-7\\6^{b}&=5\\f(6^{b}-7)&=6^{3b}-2\cdot 6^{2b}-4\\&=(6^{b})^{3}-2\cdot (6^{b})^{2}-4\\f(5-7)&=5^{3}-2\cdot 5^{2}-4\\f(-2)&=125-50-4\\&=71\\*cara2\\{\text{misalkan }}6^{b}-7=a{\text{ maka }}6^{b}=a+7\\f(6^{b}-7)&=6^{3b}-2\cdot 6^{2b}-4\\&=(6^{b})^{3}-2\cdot (6^{b})^{2}-4\\f(a)&=(a+7)^{3}-2(a+7)^{2}-4\\f(-2)&=(-2+7)^{3}-2(-2+7)^{2}-4\\&=5^{3}-2(5)^{2}-4\\&=125-50-4\\&=71\\\end{aligned}}}
Jika
f
(
x
y
)
=
f
(
x
)
y
{\displaystyle f(xy)={\frac {f(x)}{y}}}
Jawaban
f
(
10
)
=
7
f
(
2
⋅
5
)
=
7
f
(
x
y
)
=
f
(
x
)
y
f
(
2
⋅
5
)
=
f
(
2
)
5
7
=
f
(
2
)
5
f
(
2
)
=
35
{\displaystyle {\begin{aligned}f(10)&=7\\f(2\cdot 5)&=7\\f(xy)&={\frac {f(x)}{y}}\\f(2\cdot 5)&={\frac {f(2)}{5}}\\7&={\frac {f(2)}{5}}\\f(2)&=35\\\end{aligned}}}
Jika
f
(
x
y
)
=
f
(
x
+
y
)
x
y
{\displaystyle f(xy)={\frac {f(x+y)}{xy}}}
Jawaban
f
(
15
)
=
16
f
(
3
⋅
5
)
=
16
f
(
x
y
)
=
f
(
x
+
y
)
x
y
f
(
3
⋅
5
)
=
f
(
3
+
5
)
3
⋅
5
f
(
15
)
=
f
(
8
)
15
16
=
f
(
8
)
15
f
(
8
)
=
240
{\displaystyle {\begin{aligned}f(15)&=16\\f(3\cdot 5)&=16\\f(xy)&={\frac {f(x+y)}{xy}}\\f(3\cdot 5)&={\frac {f(3+5)}{3\cdot 5}}\\f(15)&={\frac {f(8)}{15}}\\16&={\frac {f(8)}{15}}\\f(8)&=240\\\end{aligned}}}
Jika
f
(
x
+
1
x
+
6
)
=
x
2
+
1
x
2
+
15
{\displaystyle f(x+{\frac {1}{x}}+6)=x^{2}+{\frac {1}{x^{2}}}+15}
Jawaban
f
(
x
+
1
x
+
6
)
=
x
2
+
1
x
2
+
15
=
(
x
+
1
x
)
2
−
2
+
15
=
(
x
+
1
x
)
2
+
13
misalkan
x
+
1
x
=
p
f
(
x
+
1
x
+
6
)
=
(
x
+
1
x
)
2
+
13
f
(
p
+
6
)
=
p
2
+
13
jika f(16) maka p adalah 10 sebelum ditambahkan 6
f
(
p
+
6
)
=
p
2
+
13
f
(
10
+
6
)
=
10
2
+
13
f
(
16
)
=
100
+
13
=
113
{\displaystyle {\begin{aligned}f(x+{\frac {1}{x}}+6)&=x^{2}+{\frac {1}{x^{2}}}+15\\&=(x+{\frac {1}{x}})^{2}-2+15\\&=(x+{\frac {1}{x}})^{2}+13\\{\text{misalkan }}x+{\frac {1}{x}}&=p\\f(x+{\frac {1}{x}}+6)&=(x+{\frac {1}{x}})^{2}+13\\f(p+6)&=p^{2}+13\\{\text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 }}\\f(p+6)&=p^{2}+13\\f(10+6)&=10^{2}+13\\f(16)&=100+13\\&=113\\\end{aligned}}}
tentukan nilai x jika
f
(
x
)
=
4
4
−
x
{\displaystyle f(x)={\frac {4}{4-x}}}
f
(
x
⋅
f
(
x
)
)
f
(
4
x
)
f
(
x
)
=
256
{\displaystyle f(x\cdot f(x))^{\frac {f(4x)}{f(x)}}=256}
Jawaban
f
(
x
)
=
4
4
−
x
f
(
4
x
)
=
4
4
−
4
x
f
(
4
x
)
f
(
x
)
=
4
4
−
4
x
4
4
−
x
=
4
−
x
4
−
4
x
f
(
x
⋅
f
(
x
)
)
=
f
(
x
(
4
4
−
x
)
)
=
f
(
4
x
4
−
x
)
=
4
4
−
(
4
x
4
−
x
)
=
4
16
−
4
x
−
4
x
4
−
x
=
4
16
−
8
x
4
−
x
=
4
(
4
−
x
)
4
(
4
−
4
x
)
=
4
−
x
4
−
4
x
misalkan
4
−
x
4
−
4
x
=
a
f
(
x
⋅
f
(
x
)
)
f
(
4
x
)
f
(
x
)
=
256
a
a
=
256
a
a
=
4
4
a
=
4
4
−
x
4
−
4
x
=
4
4
−
x
=
16
−
16
x
15
x
=
12
x
=
4
5
{\displaystyle {\begin{aligned}f(x)&={\frac {4}{4-x}}\\f(4x)&={\frac {4}{4-4x}}\\{\frac {f(4x)}{f(x)}}&={\frac {\frac {4}{4-4x}}{\frac {4}{4-x}}}\\&={\frac {4-x}{4-4x}}\\f(x\cdot f(x))&=f(x({\frac {4}{4-x}}))\\&=f({\frac {4x}{4-x}})\\&={\frac {4}{4-({\frac {4x}{4-x}})}}\\&={\frac {4}{\frac {16-4x-4x}{4-x}}}\\&={\frac {4}{\frac {16-8x}{4-x}}}\\&={\frac {4(4-x)}{4(4-4x)}}\\&={\frac {4-x}{4-4x}}\\{\text{misalkan }}{\frac {4-x}{4-4x}}&=a\\f(x\cdot f(x))^{\frac {f(4x)}{f(x)}}&=256\\a^{a}&=256\\a^{a}&=4^{4}\\a&=4\\{\frac {4-x}{4-4x}}&=4\\4-x&=16-16x\\15x&=12\\x&={\frac {4}{5}}\\\end{aligned}}}
Fungsi
f
(
x
)
=
k
x
2
x
+
1
dengan
x
≠
−
1
2
{\displaystyle f(x)={\frac {kx}{2x+1}}{\text{dengan }}x\neq -{\frac {1}{2}}}
Jawaban
f
(
x
)
=
k
x
2
x
+
1
f
(
f
(
x
)
)
=
x
f
(
k
x
2
x
+
1
)
=
x
k
(
k
x
2
x
+
1
)
2
(
k
x
2
x
+
1
)
+
1
=
x
k
2
x
2
x
+
1
2
k
x
+
2
x
+
1
2
x
+
1
=
x
k
2
x
2
k
x
+
2
x
+
1
=
x
k
2
2
k
x
+
2
x
+
1
=
1
k
2
=
2
k
x
+
2
x
+
1
k
2
−
2
k
x
=
2
x
+
1
k
2
−
2
k
x
+
x
2
=
x
2
+
2
x
+
1
(
k
−
x
)
2
=
(
x
+
1
)
2
(
k
−
x
)
2
−
(
x
+
1
)
2
=
0
(
k
−
x
+
x
+
1
)
(
k
−
x
−
(
x
+
1
)
)
=
0
k
=
−
1
atau
k
=
2
x
+
1
(TM)
{\displaystyle {\begin{aligned}f(x)&={\frac {kx}{2x+1}}\\f(f(x))&=x\\f({\frac {kx}{2x+1}})&=x\\{\frac {k({\frac {kx}{2x+1}})}{2({\frac {kx}{2x+1}})+1}}&=x\\{\frac {\frac {k^{2}x}{2x+1}}{\frac {2kx+2x+1}{2x+1}}}&=x\\{\frac {k^{2}x}{2kx+2x+1}}&=x\\{\frac {k^{2}}{2kx+2x+1}}&=1\\k^{2}&=2kx+2x+1\\k^{2}-2kx&=2x+1\\k^{2}-2kx+x^{2}&=x^{2}+2x+1\\(k-x)^{2}&=(x+1)^{2}\\(k-x)^{2}-(x+1)^{2}&=0\\(k-x+x+1)(k-x-(x+1))&=0\\k=-1&{\text{ atau }}k=2x+1&{\text{ (TM) }}\\\end{aligned}}}
Jika n = 20232 +20242 maka berapa hasil dari
2
n
−
1
{\displaystyle {\sqrt {2n-1}}}
Jawaban
n
=
2023
2
+
2024
2
=
2023
2
+
(
2023
+
1
)
2
Misalkan 2023 = p
n
=
p
2
+
(
p
+
1
)
2
=
p
2
+
p
2
+
2
p
+
1
=
2
p
2
+
2
p
+
1
2
n
−
1
=
2
(
2
p
2
+
2
p
+
1
)
−
1
=
4
p
2
+
4
p
+
2
−
1
=
4
p
2
+
4
p
+
1
=
(
2
p
+
1
)
2
=
2
p
+
1
=
2
(
2023
)
+
1
=
4046
+
1
=
4047
{\displaystyle {\begin{aligned}n&=2023^{2}+2024^{2}\\&=2023^{2}+(2023+1)^{2}\\{\text{Misalkan 2023 = p}}\\n&=p^{2}+(p+1)^{2}\\&=p^{2}+p^{2}+2p+1\\&=2p^{2}+2p+1\\{\sqrt {2n-1}}&={\sqrt {2(2p^{2}+2p+1)-1}}\\&={\sqrt {4p^{2}+4p+2-1}}\\&={\sqrt {4p^{2}+4p+1}}\\&={\sqrt {(2p+1)^{2}}}\\&=2p+1\\&=2(2023)+1\\&=4046+1\\&=4047\\\end{aligned}}}
tentukan nilai dari (a-c)b jika
a
b
a
+
b
=
1
3
{\displaystyle {\frac {ab}{a+b}}={\frac {1}{3}}}
b
c
b
+
c
=
1
4
{\displaystyle {\frac {bc}{b+c}}={\frac {1}{4}}}
a
c
a
+
c
=
1
9
{\displaystyle {\frac {ac}{a+c}}={\frac {1}{9}}}
Jawaban
a
b
a
+
b
=
1
3
a
+
b
a
b
=
3
(terbalik posisinya)
1
b
+
1
a
=
3
b
c
b
+
c
=
1
4
b
+
c
b
c
=
4
(terbalik posisinya)
1
c
+
1
b
=
4
a
c
a
+
c
=
1
9
a
+
c
a
c
=
9
(terbalik posisinya)
1
c
+
1
a
=
9
Misalkan 1/a = x, 1/b = y dan 1/c = z
x
+
y
=
3
y
+
z
=
4
x
+
z
=
9
x
+
y
=
3
y
+
z
=
4
x
−
z
=
−
1
x
−
z
=
−
1
x
+
z
=
9
2
x
=
8
x
=
4
x
−
z
=
−
1
4
−
z
=
−
1
z
=
5
x
+
y
=
3
4
+
y
=
3
y
=
−
1
1
a
=
4
a
=
1
4
1
b
=
−
1
b
=
−
1
1
c
=
5
c
=
1
5
(
a
−
c
)
b
=
(
1
4
−
1
5
)
−
1
=
(
5
−
4
20
)
−
1
=
(
1
20
)
−
1
=
20
{\displaystyle {\begin{aligned}{\frac {ab}{a+b}}&={\frac {1}{3}}\\{\frac {a+b}{ab}}&=3{\text{ (terbalik posisinya)}}\\{\frac {1}{b}}+{\frac {1}{a}}&=3\\{\frac {bc}{b+c}}&={\frac {1}{4}}\\{\frac {b+c}{bc}}&=4{\text{ (terbalik posisinya)}}\\{\frac {1}{c}}+{\frac {1}{b}}&=4\\{\frac {ac}{a+c}}&={\frac {1}{9}}\\{\frac {a+c}{ac}}&=9{\text{ (terbalik posisinya)}}\\{\frac {1}{c}}+{\frac {1}{a}}&=9\\{\text{Misalkan 1/a = x, 1/b = y dan 1/c = z}}\\x+y&=3\\y+z&=4\\x+z&=9\\x+y&=3\\y+z&=4\\x-z&=-1\\x-z&=-1\\x+z&=9\\2x&=8\\x&=4\\x-z&=-1\\4-z&=-1\\z&=5\\x+y&=3\\4+y&=3\\y&=-1\\{\frac {1}{a}}&=4\\a&={\frac {1}{4}}\\{\frac {1}{b}}&=-1\\b&=-1\\{\frac {1}{c}}&=5\\c&={\frac {1}{5}}\\(a-c)^{b}&=({\frac {1}{4}}-{\frac {1}{5}})^{-1}\\&=({\frac {5-4}{20}})^{-1}\\&=({\frac {1}{20}})^{-1}\\&=20\\\end{aligned}}}
tentukan nilai dari a, b dan c jika
a
+
b
2
=
a
+
c
4
=
b
+
c
5
{\displaystyle {\frac {a+b}{2}}={\frac {a+c}{4}}={\frac {b+c}{5}}}
Jawaban
misalkan k untuk semua ketiga persamaan tersebut
a
+
b
2
=
a
+
c
4
=
b
+
c
5
=
k
a
+
b
=
2
k
a
+
c
=
4
k
b
+
c
=
5
k
2
a
+
b
+
c
=
6
k
2
a
+
5
k
=
6
k
k
=
2
a
a
=
k
2
b
=
3
k
2
c
=
7
k
2
a
+
2
b
+
3
c
=
28
k
2
+
2
(
3
k
2
)
+
3
(
7
k
2
)
=
28
k
+
6
k
+
21
k
=
56
28
k
=
56
k
=
2
a
=
k
2
=
2
2
=
1
b
=
3
k
2
=
3
(
2
)
2
=
3
c
=
7
k
2
=
7
(
2
)
2
=
7
{\displaystyle {\begin{aligned}{\text{misalkan k untuk semua ketiga persamaan tersebut }}\\{\frac {a+b}{2}}={\frac {a+c}{4}}={\frac {b+c}{5}}&=k\\a+b&=2k\\a+c&=4k\\b+c&=5k\\2a+b+c&=6k\\2a+5k&=6k\\k&=2a\\a&={\frac {k}{2}}\\b&={\frac {3k}{2}}\\c&={\frac {7k}{2}}\\a+2b+3c&=28\\{\frac {k}{2}}+2({\frac {3k}{2}})+3({\frac {7k}{2}})&=28\\k+6k+21k&=56\\28k&=56\\k&=2\\a&={\frac {k}{2}}\\&={\frac {2}{2}}=1\\b&={\frac {3k}{2}}\\&={\frac {3(2)}{2}}=3\\c&={\frac {7k}{2}}\\&={\frac {7(2)}{2}}=7\\\end{aligned}}}
tentukan nilai dari (b+c)a jika
a
+
b
+
c
2
=
a
−
2
+
b
−
1
+
c
{\displaystyle {\frac {a+b+c}{2}}={\sqrt {a-2}}+{\sqrt {b-1}}+{\sqrt {c}}}
Jawaban
a
+
b
+
c
2
=
a
−
2
+
b
−
1
+
c
a
+
b
+
c
=
2
(
a
−
2
+
b
−
1
+
c
)
a
−
2
a
−
2
+
b
−
2
b
−
1
+
c
−
2
c
=
0
a
−
2
−
2
a
−
2
+
1
+
b
−
1
−
2
b
−
1
+
1
+
c
−
2
c
+
1
=
0
(
a
−
2
−
1
)
2
+
(
b
−
1
−
1
)
2
+
(
c
−
1
)
2
=
0
(
a
−
2
−
1
)
2
=
0
a
−
2
−
1
=
0
a
−
2
=
1
a
−
2
=
1
a
=
3
(
b
−
1
−
1
)
2
=
0
b
−
1
−
1
=
0
b
−
1
=
1
b
−
1
=
1
b
=
1
(
c
−
1
)
2
=
0
c
−
1
=
0
c
=
1
c
=
1
(
b
+
c
)
a
=
(
2
+
1
)
3
=
3
3
=
27
{\displaystyle {\begin{aligned}{\frac {a+b+c}{2}}&={\sqrt {a-2}}+{\sqrt {b-1}}+{\sqrt {c}}\\a+b+c&=2({\sqrt {a-2}}+{\sqrt {b-1}}+{\sqrt {c}})\\a-2{\sqrt {a-2}}+b-2{\sqrt {b-1}}+c-2{\sqrt {c}}&=0\\a-2-2{\sqrt {a-2}}+1+b-1-2{\sqrt {b-1}}+1+c-2{\sqrt {c}}+1&=0\\({\sqrt {a-2}}-1)^{2}+({\sqrt {b-1}}-1)^{2}+({\sqrt {c}}-1)^{2}&=0\\({\sqrt {a-2}}-1)^{2}&=0\\{\sqrt {a-2}}-1&=0\\{\sqrt {a-2}}&=1\\a-2&=1\\a&=3\\({\sqrt {b-1}}-1)^{2}&=0\\{\sqrt {b-1}}-1&=0\\{\sqrt {b-1}}&=1\\b-1&=1\\b&=1\\({\sqrt {c}}-1)^{2}&=0\\{\sqrt {c}}-1&=0\\{\sqrt {c}}&=1\\c&=1\\(b+c)^{a}&=(2+1)^{3}\\&=3^{3}\\&=27\\\end{aligned}}}
x dan y merupakan bilangan tak nol. Jika xy =
x
y
{\displaystyle {\frac {x}{y}}}
Jawaban
x
y
=
x
y
y
2
=
1
y
2
−
1
=
0
(
y
−
1
)
(
y
+
1
)
=
0
y
=
1
atau
y
=
−
1
x
y
=
x
−
y
x
=
x
y
−
y
2
x
−
x
y
=
−
y
2
x
(
1
−
y
)
=
−
y
2
x
=
−
y
2
1
−
y
cek y=1
x
=
−
1
2
1
−
1
tidak memenuhi syarat
cek y=-1
x
=
−
(
−
1
)
2
1
−
(
−
1
)
=
−
1
2
x
+
y
=
−
1
−
1
2
=
−
3
2
{\displaystyle {\begin{aligned}xy&={\frac {x}{y}}\\y^{2}&=1\\y^{2}-1&=0\\(y-1)(y+1)&=0\\y=1&{\text{ atau }}y=-1\\{\frac {x}{y}}&=x-y\\x&=xy-y^{2}\\x-xy&=-y^{2}\\x(1-y)&=-y^{2}\\x&={\frac {-y^{2}}{1-y}}\\{\text{cek y=1 }}\\x&={\frac {-1^{2}}{1-1}}\\{\text{tidak memenuhi syarat }}\\{\text{cek y=-1 }}\\x&={\frac {-(-1)^{2}}{1-(-1)}}\\&={\frac {-1}{2}}\\x+y&=-1-{\frac {1}{2}}\\&=-{\frac {3}{2}}\\\end{aligned}}}
berapa nilai x dari
(
a
b
)
3
+
(
b
a
)
3
=
2
x
{\displaystyle ({\frac {a}{b}})^{3}+({\frac {b}{a}})^{3}=2{\sqrt {x}}}
1
a
+
1
b
=
1
a
+
b
{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}={\frac {1}{a+b}}}
Jawaban
1
a
+
1
b
=
1
a
+
b
a
+
b
a
b
=
1
a
+
b
(
a
+
b
)
2
=
a
b
a
2
+
2
a
b
+
b
2
=
a
b
a
2
+
b
2
=
−
a
b
misalkan
a
b
+
b
a
=
n
a
b
+
b
a
=
n
a
2
+
b
2
a
b
=
n
a
2
+
b
2
=
n
a
b
n
=
−
1
a
b
+
b
a
=
n
(
a
b
)
3
+
(
b
a
)
3
+
3
(
a
b
+
b
a
)
=
n
3
(
a
b
)
3
+
(
b
a
)
3
+
3
n
=
n
3
(
a
b
)
3
+
(
b
a
)
3
=
n
3
−
3
n
=
(
−
1
)
3
−
3
(
−
1
)
=
2
2
x
=
2
x
=
1
x
=
1
{\displaystyle {\begin{aligned}{\frac {1}{a}}+{\frac {1}{b}}&={\frac {1}{a+b}}\\{\frac {a+b}{ab}}&={\frac {1}{a+b}}\\(a+b)^{2}&=ab\\a^{2}+2ab+b^{2}&=ab\\a^{2}+b^{2}&=-ab\\{\text{misalkan }}{\frac {a}{b}}+{\frac {b}{a}}=n\\{\frac {a}{b}}+{\frac {b}{a}}&=n\\{\frac {a^{2}+b^{2}}{ab}}&=n\\a^{2}+b^{2}&=nab\\n&=-1\\{\frac {a}{b}}+{\frac {b}{a}}&=n\\({\frac {a}{b}})^{3}+({\frac {b}{a}})^{3}+3({\frac {a}{b}}+{\frac {b}{a}})&=n^{3}\\({\frac {a}{b}})^{3}+({\frac {b}{a}})^{3}+3n&=n^{3}\\({\frac {a}{b}})^{3}+({\frac {b}{a}})^{3}&=n^{3}-3n\\&=(-1)^{3}-3(-1)\\&=2\\2{\sqrt {x}}&=2\\{\sqrt {x}}&=1\\x&=1\\\end{aligned}}}
berapa nilai m dari
x
2
−
m
x
−
1
=
0
{\displaystyle x^{2}-mx-1=0}
x
1
3
+
x
2
3
=
1
{\displaystyle {\sqrt[{3}]{x_{1}}}+{\sqrt[{3}]{x_{2}}}=1}
Jawaban
x
1
3
=
a
x
1
=
a
3
x
2
3
=
b
x
2
=
b
3
x
1
3
+
x
2
3
=
1
a
+
b
=
1
x
2
−
m
x
−
1
=
0
x
1
+
x
2
=
m
x
1
⋅
x
2
=
−
1
x
1
+
x
2
=
m
a
3
+
b
3
=
m
x
1
⋅
x
2
=
−
1
a
3
⋅
b
3
=
−
1
(
a
b
)
2
=
(
−
1
)
3
a
b
=
−
1
(
a
+
b
)
3
=
a
3
+
b
3
+
3
a
b
(
a
+
b
)
(
1
)
3
=
m
+
3
(
−
1
)
(
1
)
1
=
m
−
3
m
=
4
{\displaystyle {\begin{aligned}{\sqrt[{3}]{x_{1}}}&=a\\x_{1}&=a^{3}\\{\sqrt[{3}]{x_{2}}}&=b\\x_{2}&=b^{3}\\{\sqrt[{3}]{x_{1}}}+{\sqrt[{3}]{x_{2}}}&=1\\a+b&=1\\x^{2}-mx-1&=0\\x_{1}+x_{2}&=m\\x_{1}\cdot x_{2}&=-1\\x_{1}+x_{2}&=m\\a^{3}+b^{3}&=m\\x_{1}\cdot x_{2}&=-1\\a^{3}\cdot b^{3}&=-1\\(ab)^{2}&=(-1)^{3}\\ab&=-1\\(a+b)^{3}&=a^{3}+b^{3}+3ab(a+b)\\(1)^{3}&=m+3(-1)(1)\\1&=m-3\\m&=4\\\end{aligned}}}
berapa nilai
x
1
x
2
{\displaystyle {\frac {x_{1}}{x_{2}}}}
a
x
2
−
18
x
−
b
=
0
{\displaystyle ax^{2}-18x-b=0}
a
b
=
45
{\displaystyle ab=45}
Jawaban
a
b
=
45
b
=
45
a
a
x
2
−
18
x
−
b
=
0
a
x
2
−
18
x
−
45
a
=
0
a
2
x
2
−
18
a
x
−
45
=
0
(
a
x
−
3
)
(
a
x
−
15
)
=
0
a
x
−
3
=
0
x
=
3
a
a
x
−
15
=
0
x
=
15
a
x
1
x
2
=
3
a
15
a
=
3
15
=
1
5
x
1
x
2
=
15
a
3
a
=
15
3
=
5
{\displaystyle {\begin{aligned}ab&=45\\b&={\frac {45}{a}}\\ax^{2}-18x-b&=0\\ax^{2}-18x-{\frac {45}{a}}&=0\\a^{2}x^{2}-18ax-45&=0\\(ax-3)(ax-15)&=0\\ax-3&=0\\x&={\frac {3}{a}}\\ax-15&=0\\x&={\frac {15}{a}}\\{\frac {x_{1}}{x_{2}}}&={\frac {\frac {3}{a}}{\frac {15}{a}}}\\&={\frac {3}{15}}\\&={\frac {1}{5}}\\{\frac {x_{1}}{x_{2}}}&={\frac {\frac {15}{a}}{\frac {3}{a}}}\\&={\frac {15}{3}}\\&=5\\\end{aligned}}}
Jika
u
3
u
1
+
u
2
=
7
8
{\displaystyle {\frac {u_{3}}{u_{1}+u_{2}}}={\frac {7}{8}}}
u
2
+
u
3
u
1
{\displaystyle {\frac {u_{2}+u_{3}}{u_{1}}}}
Jawaban
u
3
u
1
+
u
2
=
7
8
a
+
2
b
a
+
a
+
b
=
7
8
a
+
2
b
2
a
+
b
=
7
8
8
(
a
+
2
b
)
=
7
(
2
a
+
b
)
8
a
+
16
b
=
14
a
+
7
b
9
b
=
6
a
b
=
2
a
3
u
2
+
u
3
u
1
=
a
+
b
+
a
+
2
b
a
=
2
a
+
3
b
a
=
2
a
+
3
(
2
a
3
)
a
=
2
a
+
2
a
a
=
4
a
a
=
4
{\displaystyle {\begin{aligned}{\frac {u_{3}}{u_{1}+u_{2}}}&={\frac {7}{8}}\\{\frac {a+2b}{a+a+b}}&={\frac {7}{8}}\\{\frac {a+2b}{2a+b}}&={\frac {7}{8}}\\8(a+2b)&=7(2a+b)\\8a+16b&=14a+7b\\9b&=6a\\b&={\frac {2a}{3}}\\{\frac {u_{2}+u_{3}}{u_{1}}}&={\frac {a+b+a+2b}{a}}\\&={\frac {2a+3b}{a}}\\&={\frac {2a+3({\frac {2a}{3}})}{a}}\\&={\frac {2a+2a}{a}}\\&={\frac {4a}{a}}\\&=4\\\end{aligned}}}
Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya? Jawaban
7
p
+
q
2
p
+
q
=
17
p
+
q
7
p
+
q
(
7
p
+
q
)
2
=
(
17
p
+
q
)
(
2
p
+
q
)
49
p
2
+
14
p
q
+
q
2
=
34
p
2
+
19
p
q
+
q
2
15
p
2
=
5
p
q
3
p
=
q
7
p
+
q
2
p
+
q
=
7
p
+
3
p
2
p
+
3
p
=
10
p
5
p
=
2
{\displaystyle {\begin{aligned}{\frac {7p+q}{2p+q}}&={\frac {17p+q}{7p+q}}\\(7p+q)^{2}&=(17p+q)(2p+q)\\49p^{2}+14pq+q^{2}&=34p^{2}+19pq+q^{2}\\15p^{2}&=5pq\\3p&=q\\{\frac {7p+q}{2p+q}}&={\frac {7p+3p}{2p+3p}}\\&={\frac {10p}{5p}}\\&=2\\\end{aligned}}}
Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b? Jawaban
rataan geometris
a
⋅
b
=
b
−
24
a
⋅
b
=
(
b
−
24
)
2
rataan aritmatik
a
+
b
2
=
a
+
15
a
+
b
=
2
(
a
+
15
)
a
+
b
=
2
a
+
30
a
=
b
−
30
a
⋅
b
=
(
b
−
24
)
2
(
b
−
30
)
b
=
(
b
−
24
)
2
b
2
−
30
b
=
b
2
−
48
b
+
576
18
b
=
576
b
=
32
a
=
b
−
30
=
32
−
30
=
2
a
+
b
=
32
+
2
=
34
{\displaystyle {\begin{aligned}{\text{rataan geometris }}\\{\sqrt {a\cdot b}}&=b-24\\a\cdot b&=(b-24)^{2}\\{\text{rataan aritmatik }}\\{\frac {a+b}{2}}&=a+15\\a+b&=2(a+15)\\a+b&=2a+30\\a&=b-30\\a\cdot b&=(b-24)^{2}\\(b-30)b&=(b-24)^{2}\\b^{2}-30b&=b^{2}-48b+576\\18b&=576\\b&=32\\a&=b-30\\&=32-30\\&=2\\a+b&=32+2\\&=34\\\end{aligned}}}
Segitiga lancip ABC dengan
a
4
+
b
4
+
c
4
+
a
2
b
2
c
2
(
a
2
+
b
2
)
=
2
{\displaystyle {\frac {a^{4}+b^{4}+c^{4}+a^{2}b^{2}}{c^{2}(a^{2}+b^{2})}}=2}
Jawaban
syarat segitiga lancip semua sudut masing-masing kurang dari
90
∘
c
2
=
a
2
+
b
2
−
2
a
b
c
o
s
C
c
o
s
C
=
a
2
+
b
2
−
c
2
2
a
b
a
4
+
b
4
+
c
4
+
a
2
b
2
=
2
c
2
(
a
2
+
b
2
)
a
4
+
b
4
+
a
2
b
2
+
c
4
=
2
c
2
(
a
2
+
b
2
)
(
a
2
+
b
2
)
2
−
a
2
b
2
+
c
4
=
2
c
2
(
a
2
+
b
2
)
(
a
2
+
b
2
)
2
−
2
c
2
(
a
2
+
b
2
)
+
(
c
2
)
2
=
a
2
b
2
(
a
2
+
b
2
−
c
2
)
2
=
a
2
b
2
(
a
2
+
b
2
−
c
2
)
2
=
(
a
b
)
2
a
2
+
b
2
−
c
2
=
±
a
b
c
o
s
C
=
±
a
b
2
a
b
=
±
1
2
=
1
2
(karena sudut harus kurang dari
90
∘
)
C
=
60
∘
{\displaystyle {\begin{aligned}{\text{syarat segitiga lancip semua sudut masing-masing kurang dari }}90^{\circ }\\c^{2}&=a^{2}+b^{2}-2abcosC\\cosC&={\frac {a^{2}+b^{2}-c^{2}}{2ab}}\\a^{4}+b^{4}+c^{4}+a^{2}b^{2}&=2c^{2}(a^{2}+b^{2})\\a^{4}+b^{4}+a^{2}b^{2}+c^{4}&=2c^{2}(a^{2}+b^{2})\\(a^{2}+b^{2})^{2}-a^{2}b^{2}+c^{4}&=2c^{2}(a^{2}+b^{2})\\(a^{2}+b^{2})^{2}-2c^{2}(a^{2}+b^{2})+(c^{2})^{2}&=a^{2}b^{2}\\(a^{2}+b^{2}-c^{2})^{2}&=a^{2}b^{2}\\(a^{2}+b^{2}-c^{2})^{2}&=(ab)^{2}\\a^{2}+b^{2}-c^{2}&=\pm ab\\cosC&=\pm {\frac {ab}{2ab}}\\&=\pm {\frac {1}{2}}\\&={\frac {1}{2}}{\text{ (karena sudut harus kurang dari }}90^{\circ })\\C&=60^{\circ }\\\end{aligned}}}
Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm2 ? Jawaban
misalkan panjang DA dan EA
=
x
dan panjang AB
=
y
luas segitiga CAB
=
C
A
⋅
A
B
2
45
=
(
x
+
9
)
(
x
+
5
)
2
90
=
x
2
+
14
x
+
45
x
2
+
14
x
=
45
y
2
=
(
x
+
9
)
2
+
(
x
+
5
)
2
=
x
2
+
18
x
+
81
+
x
2
+
10
x
+
25
=
2
x
2
+
28
x
+
106
=
2
(
x
2
+
14
x
)
+
106
=
2
(
45
)
+
106
=
196
y
=
14
{\displaystyle {\begin{aligned}{\text{misalkan panjang DA dan EA }}=x{\text{ dan panjang AB }}=y\\{\text{luas segitiga CAB }}&={\frac {CA\cdot AB}{2}}\\45&={\frac {(x+9)(x+5)}{2}}\\90&=x^{2}+14x+45\\x^{2}+14x&=45\\y^{2}&=(x+9)^{2}+(x+5)^{2}\\&=x^{2}+18x+81+x^{2}+10x+25\\&=2x^{2}+28x+106\\&=2(x^{2}+14x)+106\\&=2(45)+106\\&=196\\y&=14\\\end{aligned}}}
jadi panjang BC adalah 14 cm
Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG? Jawaban
kita cari ukuran GC yaitu
G
C
A
D
=
C
E
D
E
G
C
15
=
6
18
G
C
=
5
luas BEFG = luas segitiga BFC - luas segitiga GEC
=
1
2
⋅
B
C
⋅
C
F
−
1
2
⋅
G
C
⋅
C
E
=
1
2
⋅
15
⋅
18
−
1
2
⋅
5
⋅
6
=
135
−
15
=
120
{\displaystyle {\begin{aligned}{\text{kita cari ukuran GC yaitu }}\\{\frac {GC}{AD}}&={\frac {CE}{DE}}\\{\frac {GC}{15}}&={\frac {6}{18}}\\GC&=5\\{\text{luas BEFG = luas segitiga BFC - luas segitiga GEC }}\\&={\frac {1}{2}}\cdot BC\cdot CF-{\frac {1}{2}}\cdot GC\cdot CE\\&={\frac {1}{2}}\cdot 15\cdot 18-{\frac {1}{2}}\cdot 5\cdot 6\\&=135-15\\&=120\\\end{aligned}}}
jadi luas daerah BFEG adalah 120 cm2
Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD? Jawaban
A
I
=
1
5
A
F
=
1
5
10
=
2
I
F
=
A
F
−
A
I
=
10
−
2
=
8
luas trapesium AFGD
=
(
A
D
+
E
F
)
⋅
A
F
2
=
(
4
+
6
)
10
2
=
50
luas segitiga AID
=
A
I
⋅
A
F
2
=
(
2
)
4
2
=
4
luas segitiga IFG
=
I
F
⋅
F
G
2
=
(
8
)
6
2
=
24
luas daerah segitiga IGD
=
luas trapesium AFGD-luas segitiga AI—luas segitiga IFG
=
50
−
4
−
24
=
22
{\displaystyle {\begin{aligned}\\AI&={\frac {1}{5}}AF\\&={\frac {1}{5}}10\\&=2\\IF&=AF-AI\\&=10-2\\&=8\\{\text{luas trapesium AFGD }}&={\frac {(AD+EF)\cdot AF}{2}}\\&={\frac {(4+6)10}{2}}\\&=50\\{\text{luas segitiga AID }}&={\frac {AI\cdot AF}{2}}\\&={\frac {(2)4}{2}}\\&=4\\{\text{luas segitiga IFG }}&={\frac {IF\cdot FG}{2}}\\&={\frac {(8)6}{2}}\\&=24\\{\text{luas daerah segitiga IGD }}&={\text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG }}\\&=50-4-24\\&=22\\\end{aligned}}}
jadi luas daerah segitiga IGD adalah 22 cm2
Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-
π
{\displaystyle \pi }
3 maka berapa luas permukaan kerucut tersebut? Jawaban
volume balok
V
b
=
x
2
(
12
)
volume kerucut
V
b
=
1
3
π
x
2
(
12
)
=
4
π
x
2
V
b
−
k
=
V
b
−
V
k
100
(
3
−
π
)
=
12
x
2
−
4
π
x
2
100
(
3
−
π
)
=
4
x
2
(
3
−
π
)
x
2
=
25
x
=
5
s
=
12
2
+
5
2
=
144
+
25
=
169
=
13
luas permukaan kerucut
=
π
r
(
r
+
s
)
=
π
(
5
)
(
5
+
13
)
=
90
π
{\displaystyle {\begin{aligned}\\{\text{volume balok}}\\V_{b}&=x^{2}(12)\\{\text{volume kerucut}}\\V_{b}&={\frac {1}{3}}\pi x^{2}(12)\\&=4\pi x^{2}\\V_{b-k}&=Vb-Vk\\100(3-\pi )&=12x^{2}-4\pi x^{2}\\100(3-\pi )&=4x^{2}(3-\pi )\\x^{2}&=25\\x&=5\\s&={\sqrt {12^{2}+5^{2}}}\\&={\sqrt {144+25}}\\&={\sqrt {169}}\\&=13\\{\text{luas permukaan kerucut }}&=\pi r(r+s)\\&=\pi (5)(5+13)\\&=90\pi \\\end{aligned}}}
jadi luas daerah permukaan kerucut adalah 90
π
{\displaystyle \pi }
2
Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9? Jawaban
A
=
3
a
+
1
B
=
3
b
+
2
A
(
A
+
1
)
+
3
B
(
3
a
+
1
)
(
3
a
+
1
+
1
)
+
3
(
3
b
+
2
)
(
3
a
+
1
)
(
3
a
+
2
)
+
9
b
+
6
9
a
2
+
9
a
+
2
+
9
b
+
6
9
a
2
+
9
a
+
9
b
+
8
9
(
a
2
+
a
+
b
)
+
8
sisa pembagiannya adalah
8
{\displaystyle {\begin{aligned}A&=3a+1\\B&=3b+2\\A(A+1)+3B\\(3a+1)(3a+1+1)+3(3b+2)\\(3a+1)(3a+2)+9b+6\\9a^{2}+9a+2+9b+6\\9a^{2}+9a+9b+8\\9(a^{2}+a+b)+8\\{\text{sisa pembagiannya adalah }}8\\\end{aligned}}}
Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81? Jawaban
A
=
9
a
+
7
B
=
9
b
+
8
A
(
A
−
5
)
+
9
B
(
9
a
+
7
)
(
9
a
+
7
−
5
)
+
9
(
9
b
+
8
)
(
9
a
+
7
)
(
9
a
+
2
)
+
81
b
+
72
81
a
2
+
81
a
+
14
+
81
b
+
72
81
a
2
+
81
a
+
81
b
+
86
81
a
2
+
81
a
+
81
b
+
81
+
5
81
(
a
2
+
a
+
b
+
1
)
+
5
sisa pembagiannya adalah
5
{\displaystyle {\begin{aligned}A&=9a+7\\B&=9b+8\\A(A-5)+9B\\(9a+7)(9a+7-5)+9(9b+8)\\(9a+7)(9a+2)+81b+72\\81a^{2}+81a+14+81b+72\\81a^{2}+81a+81b+86\\81a^{2}+81a+81b+81+5\\81(a^{2}+a+b+1)+5\\{\text{sisa pembagiannya adalah }}5\\\end{aligned}}}
Jika
[
3
7
−
1
−
2
]
{\displaystyle {\begin{bmatrix}3&7\\-1&-2\\\end{bmatrix}}}
21 +A25 +A46 ? Jawaban
A
2
=
A
⋅
A
=
[
3
7
−
1
−
2
]
⋅
[
3
7
−
1
−
2
]
=
[
2
7
−
1
−
3
]
A
3
=
A
2
⋅
A
=
[
2
7
−
1
−
3
]
⋅
[
3
7
−
1
−
2
]
=
[
−
1
0
0
−
1
]
=
−
[
1
0
0
1
]
=
−
I
A
21
+
A
25
+
A
46
=
A
21
⋅
(
I
+
A
4
+
A
25
)
=
A
21
⋅
(
I
+
A
3
⋅
A
+
A
24
⋅
A
)
=
(
A
3
)
7
⋅
(
I
+
A
3
⋅
A
+
(
A
3
)
8
⋅
A
)
=
(
−
I
)
7
⋅
(
I
−
I
⋅
A
+
(
−
I
)
8
⋅
A
)
=
−
I
⋅
(
I
−
A
+
A
)
=
−
I
⋅
I
=
−
I
=
−
[
1
0
0
1
]
=
[
−
1
0
0
−
1
]
{\displaystyle {\begin{aligned}A^{2}&=A\cdot A\\&={\begin{bmatrix}3&7\\-1&-2\\\end{bmatrix}}\cdot {\begin{bmatrix}3&7\\-1&-2\\\end{bmatrix}}={\begin{bmatrix}2&7\\-1&-3\\\end{bmatrix}}\\A^{3}&=A^{2}\cdot A\\&={\begin{bmatrix}2&7\\-1&-3\\\end{bmatrix}}\cdot {\begin{bmatrix}3&7\\-1&-2\\\end{bmatrix}}={\begin{bmatrix}-1&0\\0&-1\\\end{bmatrix}}\\&=-{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}\\&=-I\\A^{21}+A^{25}+A^{46}&=A^{21}\cdot (I+A^{4}+A^{25})\\&=A^{21}\cdot (I+A^{3}\cdot A+A^{24}\cdot A)\\&=(A^{3})^{7}\cdot (I+A^{3}\cdot A+(A^{3})^{8}\cdot A)\\&=(-I)^{7}\cdot (I-I\cdot A+(-I)^{8}\cdot A)\\&=-I\cdot (I-A+A)\\&=-I\cdot I\\&=-I\\&=-{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}\\&={\begin{bmatrix}-1&0\\0&-1\\\end{bmatrix}}\\\end{aligned}}}
Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida? bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15
ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9).
ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8)
maka Ida menulis bilangan 8. Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9? Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu.
untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja.
untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja.
Jadi banyaknya bilangan mungkin 2. Buktikan bahwa 8n dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli! cara 1 81 = 1
82 = 1 (82 =81 x81 sama dengan 1x1)
83 = 1 (83 =81 x82 sama dengan 1x1)
84 = 1 (84 =81 x83 sama dengan 1x1 atau 84 =(82 )2 sama dengan 1^2)
85 = 1
8n = 1 (semua n untuk bilangan asli) Terbukti 8n dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli
cara 2 8n = b mod 7 81 = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 81 dianggap pangkat terkecil)
(81 )n = 1n mod 7 (pangkat n kedua ruasnya)
8n = 1n mod 7
8n = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8n dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli
Berapa hasil sisa dari 1799 dibagi 5? cara 1 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 71 = 7 (sisa 1)
72 = 49 (sisa 2)
73 = 343 (sisa 3)
74 = 2,401 (sisa 0)
75 = 16,807
76 = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3
cara 2
171 = 2
172 = 4
173 = 3
174 = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3
1799 = (174 )24 x 173 Untuk 174 hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 1799 dibagi 7 sama dengan sisa 173 dibagi 7 yaitu 3. Jadi 1799 dibagi 7 bersisa 3
cara 3
Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 174
174 = 1 mod 5
(174 )24 = 124 mod 5
1796 = 124 mod 5
1796 = 1 mod 5
1796 x 173 = 1 x 173 mod 5
1799 = 173 mod 5
1799 = 17 x 17 x 17 mod 5
1799 = 2 x 2 x 2 mod 5
1799 = 8 mod 5
1799 = 3 mod 5 Jadi 1799 dibagi 5 bersisa 3
Berapa hasil sisa dari 1799 dibagi 7? cara 1
171 = 3
172 = 2
173 = 6
174 = 4
175 = 5
176 = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3
1799 = (176 )16 x 173 Untuk 176 hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 1799 dibagi 7 sama dengan sisa 173 dibagi 7 yaitu 6. Jadi 1799 dibagi 7 bersisa 6
cara 2
Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 176
176 = 1 mod 7
(176 )16 = 116 mod 7
1796 = 116 mod 7
1796 = 1 mod 7
1796 x 173 = 1 x 173 mod 7
1799 = 173 mod 7
1799 = 17 x 17 x 17 mod 7
1799 = 3 x 3 x 3 mod 7
1799 = 27 mod 7
1799 = 6 mod 7 Jadi 1799 dibagi 7 bersisa 6
Berapa hasil sisa dari 412024 dibagi 33? Jawaban
41
2024
=
41
2024
mod
33
=
(
33
×
3
+
2
)
2024
mod
33
=
2
2024
mod
33
=
2
2020
2
4
mod
33
=
(
2
5
)
404
2
4
mod
33
=
(
33
−
1
)
404
2
4
mod
33
=
(
−
1
)
404
2
4
mod
33
=
2
4
mod
33
=
16
mod
33
Jadi hasil sisa adalah
16
{\displaystyle {\begin{aligned}41^{2024}&=41^{2024}{\text{ mod }}33\\&=(33\times 3+2)^{2024}{\text{ mod }}33\\&=2^{2024}{\text{ mod }}33\\&=2^{2020}2^{4}{\text{ mod }}33\\&=(2^{5})^{404}2^{4}{\text{ mod }}33\\&=(33-1)^{404}2^{4}{\text{ mod }}33\\&=(-1)^{404}2^{4}{\text{ mod }}33\\&=2^{4}{\text{ mod }}33\\&=16{\text{ mod }}33\\{\text{Jadi hasil sisa adalah }}16\\\end{aligned}}}
Berapa nilai bilangan n terbesar sehingga 243n membagi 9999 ? Jawaban
99
99
=
(
3
2
×
11
)
99
=
3
198
×
11
99
243
n
=
(
3
5
)
n
=
3
5
n
agar bisa membagi, maka
5
n
=
198
n
=
39.6
jadi bilangan n terbesar adalah
39
{\displaystyle {\begin{aligned}99^{99}&=(3^{2}\times 11)^{99}\\&=3^{198}\times 11^{99}\\243^{n}&=(3^{5})^{n}\\&=3^{5n}\\{\text{agar bisa membagi, maka}}\\5n&=198\\n&=39.6\\{\text{jadi bilangan n terbesar adalah }}39\\\end{aligned}}}
Berapa nilai bilangan n terbesar sehingga 512n membagi 8888 ? Jawaban
88
88
=
(
8
×
11
)
88
=
8
88
×
11
88
=
8
87
×
8
×
11
88
=
(
8
3
)
29
×
8
×
11
88
=
512
29
×
8
×
11
88
512
n
=
512
29
jadi bilangan n terbesar adalah
29
{\displaystyle {\begin{aligned}88^{88}&=(8\times 11)^{88}\\&=8^{88}\times 11^{88}\\&=8^{87}\times 8\times 11^{88}\\&=(8^{3})^{29}\times 8\times 11^{88}\\&=512^{29}\times 8\times 11^{88}\\512^{n}&=512^{29}\\{\text{jadi bilangan n terbesar adalah }}29\\\end{aligned}}}
Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6! Cara 1
KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. N dibagi 3 sisa 1
N dibagi 5 sisa 2
N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a
KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70
KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21
KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97
Cara 2
Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35
kemudian ditambahkan sisa masing-masing sesuai dengan KPK. KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, 97
KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97
KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, 97 Jadi bilangan bulat positif adalah 97
NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember? Cara 1
Ember A (5 l)
Ember B (3 l)
Keterangan
5
0
Isikan 5 l ke ember A
2
3
Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2
2
0
Semua isi ember B dibuang
0
2
Tuangkan sisa ember A ke B
5
2
Isikan 5 l ke ember A
4
3
Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4
4
0
Semua isi ember B dibuang
1
3
Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1
nah ada ember A berisi 1 liter.
Cara 2
Ember A (3 l)
Ember B (5 l)
Keterangan
3
0
Isikan 3 l ke ember A
0
3
Tuangkan 3 l dari ember A ke B sehingga ember A kosong
3
3
Isikan 3 l ke ember A
1
5
Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1
nah ada ember A berisi 1 liter.
Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember? Cara 1
Ember A (5 l)
Ember B (3 l)
Keterangan
5
0
Isikan 5 l ke ember A
2
3
Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2
2
0
Semua isi ember B dibuang
0
2
Tuangkan sisa ember A ke B
5
2
Isikan 5 l ke ember A
4
3
Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4
nah ada ember A berisi 4 liter.
Cara 2
Ember A (3 l)
Ember B (5 l)
Keterangan
3
0
Isikan 3 l ke ember A
0
3
Tuangkan 3 l dari ember A ke B sehingga ember A kosong
3
3
Isikan 3 l ke ember A
1
5
Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1
1
0
Semua isi ember B dibuang
0
1
Tuangkan 1 l dari ember A ke B sehingga ember A kosong
3
1
Isikan 3 l ke ember A
0
4
Tuangkan 3 l dari ember A ke B sehingga ember A kosong
nah ada ember B berisi 4 liter.
![{\displaystyle {\frac {\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {\dots }}}}}}}}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}\dots }}}}}}}={\frac {9}{10}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/053a838885a4851eec052baabd4561705da58903)
![{\displaystyle {\begin{aligned}{\frac {\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {\dots }}}}}}}}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}\dots }}}}}}}&={\frac {9}{10}}\\{\text{misalkan untuk }}{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {\dots }}}}}}}}=p\\{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {\dots }}}}}}}}&=p\\x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {x^{2}-x-{\sqrt {\dots }}}}}}&=p^{2}\\x^{2}-x-p&=p^{2}\\x^{2}-2x+1+x-1&=p^{2}+p\\(x-1)^{2}+(x-1)&=p^{2}+p\\x-1&=p\\{\text{misalkan untuk }}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}\dots }}}}}}&=q\\{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}\dots }}}}}}&=q\\x^{2}{\sqrt[{3}]{x^{2}{\sqrt[{3}]{x^{2}\dots }}}}&=q^{3}\\x^{2}q&=q^{3}\\x^{2}&=q^{2}\\x&=q\\{\frac {x-1}{x}}&={\frac {9}{10}}\\x&=10\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/079ce3f0a97e3eade58dfb888c69da77af6317f4)
![{\displaystyle {\sqrt {1+{\sqrt {1+x}}}}={\sqrt[{3}]{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/727e3dc360730379f72685dad03843c8b17fb88a)
![{\displaystyle {\begin{aligned}{\sqrt {1+{\sqrt {1+x}}}}&={\sqrt[{3}]{x}}\\{\sqrt[{3}]{x}}&=n\\x&=n^{3}\\{\sqrt {1+{\sqrt {1+n^{3}}}}}&=n\\1+{\sqrt {1+n^{3}}}&=n^{2}\\{\sqrt {1+n^{3}}}&=n^{2}-1\\1+n^{3}&=n^{4}-2n^{2}+1\\n^{4}-n^{3}-2n^{2}&=0\\n^{2}(n^{2}-n-2)&=0\\n^{2}(n-2)(n+1)&=0\\n=0,n=2{\text{ atau }}n=-1\\n&=0\\x&=0^{3}\\&=0\\n&=2\\x&=2^{3}\\&=8\\n&=-1\\x&=(-1)^{3}\\&=-1\\{\text{yang paling mungkin untuk nilai x adalah }}8\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa1f59b672e470f104586f41fc15ad19fd73e5d8)
![{\displaystyle {\sqrt[{3}]{(8+x)^{2}}}-{\sqrt[{3}]{(8+x)(27-x)}}+{\sqrt[{3}]{(27-x)^{2}}}=7}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f10b323836c312257017e8517287fcdb16eb74d2)
![{\displaystyle {\begin{aligned}{\sqrt[{3}]{(8+x)^{2}}}-{\sqrt[{3}]{(8+x)(27-x)}}+{\sqrt[{3}]{(27-x)^{2}}}&=7\\({\sqrt[{3}]{8+x}})^{2}-{\sqrt[{3}]{8+x}}{\sqrt[{3}]{27-x}}+({\sqrt[{3}]{27-x}})^{2}&=7\\{\text{misalkan }}{\sqrt[{3}]{8+x}}=a,8+x=a^{3},{\sqrt[{3}]{27-x}}=b{\text{ dan }}27-x=b^{3}\\a^{2}-ab+b^{2}&=7\\a^{3}+b^{3}&=8+x+27-x\\&=35\\a^{3}+b^{3}&=(a+b)(a^{2}-ab+b^{2})\\35&=(a+b)(7)\\a+b&=5\\b&=5-a\\(a+b)^{3}&=a^{3}+b^{3}+3ab(a+b)\\5^{3}&=35+3ab(5)\\125&=35+15ab\\80&=15ab\\ab&=6\\a(5-a)&=6\\5a-a^{2}&=6\\a^{2}-5a+6&=6\\(a-2)(a-3)&=6\\a=2&{\text{ atau }}a=3\\a=2,b=3{\text{ dan }}a=3,b=2\\8+x&=a^{3}\\&=2^{3}\\&=8\\x&=0\\8+x&=a^{3}\\&=3^{3}\\&=27\\x&=19\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c8f9e154ac25d68ac7e10af948dd07996b01bcc)
![{\displaystyle {\sqrt[{n}]{\frac {x^{n}+4^{n}}{x^{n}+16^{n}}}}-{\frac {1}{2}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd2b3fbdbe60b362f1127933bb4b995132ca674b)
![{\displaystyle {\begin{aligned}{\sqrt[{n}]{\frac {x^{n}+4^{n}}{x^{n}+16^{n}}}}-{\frac {1}{2}}&=0\\{\sqrt[{n}]{\frac {x^{n}+4^{n}}{x^{n}+16^{n}}}}&={\frac {1}{2}}\\{\frac {x^{n}+4^{n}}{x^{n}+16^{n}}}&=({\frac {1}{2}})^{n}\\{\frac {x^{n}+4^{n}}{x^{n}+16^{n}}}&={\frac {1}{2^{n}}}\\2^{n}(x^{n}+4^{n})&=x^{n}+16^{n}\\2^{n}(x^{n}+2^{2n})&=x^{n}+2^{4n}\\2^{n}\cdot x^{n}+2^{3n}&=x^{n}+2^{4n}\\2^{n}\cdot x^{n}-x^{n}&=2^{4n}-2^{3n}\\x^{n}(2^{n}-1)&=2^{3n}(2^{n}-1)\\x^{n}&=2^{3n}\\x^{n}&=(2^{3})^{n}\\x^{n}&=8^{n}\\x&=8\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8e3cb016c307eb775252a53789d87c4adc21842)
![{\displaystyle {\sqrt[{3}]{x}}-{\sqrt[{3}]{x-36}}=3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd06f06204b35a30ba842044d58c7d5a62072569)
![{\displaystyle {\begin{aligned}{\sqrt[{3}]{x}}-{\sqrt[{3}]{x-36}}&=3\\({\sqrt[{3}]{x}}-{\sqrt[{3}]{x-36}})^{3}&=3^{3}\\x-(x-36)-3{\sqrt[{3}]{x(x-36)}}({\sqrt[{3}]{x}}-{\sqrt[{3}]{x-36}})&=27\\36-3{\sqrt[{3}]{x(x-36)}}3&=27\\-9{\sqrt[{3}]{x(x-36)}}&=-9\\{\sqrt[{3}]{x(x-36)}}&=1\\x(x-36)&=1\\x^{2}-36x-1&=0\\x-36-{\frac {1}{x}}&=0\\x-{\frac {1}{x}}&=36\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16c74aebbfcf48b7628ed32785447d7dd6a10529)
![{\displaystyle x+{\sqrt[{5}]{8}}+{\sqrt[{5}]{2}}=1+{\sqrt[{5}]{16}}+{\sqrt[{5}]{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f443375854a4e1b781befe83f7ab93a75295089)
![{\displaystyle {\begin{aligned}x+{\sqrt[{5}]{8}}+{\sqrt[{5}]{2}}&=1+{\sqrt[{5}]{16}}+{\sqrt[{5}]{4}}\\x+({\sqrt[{5}]{2}})^{3}+{\sqrt[{5}]{2}}&=1+({\sqrt[{5}]{2}})^{4}+({\sqrt[{5}]{2}})^{2}\\x&=({\sqrt[{5}]{2}})^{4}-({\sqrt[{5}]{2}})^{3}+({\sqrt[{5}]{2}})^{2}-{\sqrt[{5}]{2}}+1\\{\text{misalkan }}{\sqrt[{5}]{2}}=p\\x&=p^{4}-p^{3}+p^{2}-p+1\\x&={\frac {p^{5}+1}{p+1}}\\({\frac {x-3}{x}})^{25}&=(1-{\frac {3}{x}})^{25}\\&=(1-{\frac {3}{\frac {p^{5}+1}{p+1}}})^{25}\\&=(1-{\frac {3(p+1)}{p^{5}+1}})^{25}\\&=(1-{\frac {3({\sqrt[{5}]{2}}+1)}{({\sqrt[{5}]{2}})^{5}+1}})^{25}\\&=(1-{\frac {(3{\sqrt[{5}]{2}}+3)}{2+1}})^{25}\\&=(1-{\frac {(3{\sqrt[{5}]{2}}+3)}{3}})^{25}\\&=({\frac {3-(3{\sqrt[{5}]{2}}+3)}{3}})^{25}\\&=({\frac {3-3{\sqrt[{5}]{2}}-3)}{3}})^{25}\\&=(-{\sqrt[{5}]{2}})^{25}\\&=(-2)^{5}\\&=-32\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2420795908f45f0207b831b3670e13e45a846bc6)
![{\displaystyle {\sqrt[{3}]{x_{1}}}+{\sqrt[{3}]{x_{2}}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a28a5cf9d08ce2e8cbbe78306ecfdd83a9d3813c)
![{\displaystyle {\begin{aligned}{\sqrt[{3}]{x_{1}}}&=a\\x_{1}&=a^{3}\\{\sqrt[{3}]{x_{2}}}&=b\\x_{2}&=b^{3}\\{\sqrt[{3}]{x_{1}}}+{\sqrt[{3}]{x_{2}}}&=1\\a+b&=1\\x^{2}-mx-1&=0\\x_{1}+x_{2}&=m\\x_{1}\cdot x_{2}&=-1\\x_{1}+x_{2}&=m\\a^{3}+b^{3}&=m\\x_{1}\cdot x_{2}&=-1\\a^{3}\cdot b^{3}&=-1\\(ab)^{2}&=(-1)^{3}\\ab&=-1\\(a+b)^{3}&=a^{3}+b^{3}+3ab(a+b)\\(1)^{3}&=m+3(-1)(1)\\1&=m-3\\m&=4\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc35d185bf06fe47a6a4fc8c5a9cd3a0800ccc49)
