Dari Wikibuku bahasa Indonesia, sumber buku teks bebas
Pangkat dua/Kuadrat 1. Hitunglah!
82
422
1512 Jawaban
82 = 64
422 = 1,764
1512 = 22,801 Akar dua/Kuadrat 1. Hitunglah!
1.
81
2.
2
,
916
3.
63
,
001
{\displaystyle {\begin{aligned}&1.{\sqrt {81}}\\&2.{\sqrt {2,916}}\\&3.{\sqrt {63,001}}\\\end{aligned}}}
Jawaban
1.
81
=
9
2.
2
,
916
=
54
3.
63
,
001
=
251
{\displaystyle {\begin{aligned}&1.{\sqrt {81}}=9\\&2.{\sqrt {2,916}}=54\\&3.{\sqrt {63,001}}=251\\\end{aligned}}}
Pangkat tiga/Kubik 1. Hitunglah!
63
643
3243 Jawaban
63 = 216
643 = 262,144
3243 = 34,012,224 Akar tiga/Kubik
Pembahasan untuk mencari hasil dari akar tiga/Kubik Data:
Pangkat tiga
Hasil
Pangkat tiga
Hasil
1
1
6
216
2
8
7
343
3
27
8
512
4
64
9
729
5
125
10
1,000
Langkah-langkahnya hasil dimulainya dari angka terakhir ke angka depan:
Untuk satuan, perhatikan satuan dari hasil masing-masing sesuai dengan berurutan sebagai berikut: 1 (1), 2 (8), 3 (7), 4 (4), 5 (5), 6 (6), 7 (3), 8 (2), 9 (9) dan 0 (0). Dalam satuan tersebut maka satuan bernilai tetap adalah 0, 1, 4, 5, 6 dan 9 sedangkan satuan berubah dan posisinya terbalik adalah 2, 3, 7 dan 8.
Untuk puluhan, ratusan, dsb. Sisipan tiga basis masing-masing, tiap basis hasil angka itu antara hasil sebelumnya dan sesudahnya dan terambil angka hasil sebelumnya contoh: 412 itu berarti antara 343 dan 512 jadi terambilnya 343 berarti 7. 1. Hitunglah!
1.
64
3
2.
474
,
552
3
3.
48
,
228
,
544
3
{\displaystyle {\begin{aligned}&1.{\sqrt[{3}]{64}}\\&2.{\sqrt[{3}]{474,552}}\\&3.{\sqrt[{3}]{48,228,544}}\\\end{aligned}}}
Jawaban
1.
64
3
=
4
2.
474
,
552
3
=
78
3.
48
,
228
,
544
3
=
364
{\displaystyle {\begin{aligned}&1.{\sqrt[{3}]{64}}=4\\&2.{\sqrt[{3}]{474,552}}=78\\&3.{\sqrt[{3}]{48,228,544}}=364\\\end{aligned}}}
Angka terakhir dalam berpangkat
Pangkat 1
Pangkat 2
Pangkat 3
Pangkat 4
2
2
4
8
6
3
3
9
7
1
4
4
6
4
6
7
7
9
3
1
8
8
4
2
6
9
9
1
9
1
Keterangan
untuk pangkat 5,6,7, dst pasti berulang yang sama dengan keempat angka secara berurutan diatas data tsb.
untuk angka 0,1,5 dan 6 dimana hasil akhir angka terakhir selalu sama dengan angka satuan utamanya.
Perakaran berulang
Hasil
a
+
a
+
a
+
…
{\displaystyle {\sqrt {a+{\sqrt {a+{\sqrt {a+\dots }}}}}}}
1
+
1
+
4
a
2
{\displaystyle {\frac {1+{\sqrt {1+4a}}}{2}}}
a
−
a
−
a
−
…
{\displaystyle {\sqrt {a-{\sqrt {a-{\sqrt {a-\dots }}}}}}}
−
1
+
1
+
4
a
2
{\displaystyle {\frac {-1+{\sqrt {1+4a}}}{2}}}
a
⋅
a
⋅
a
⋅
…
{\displaystyle {\sqrt {a\cdot {\sqrt {a\cdot {\sqrt {a\cdot \dots }}}}}}}
a
a
a
a
…
{\displaystyle {\sqrt {\frac {a}{\sqrt {\frac {a}{\sqrt {\frac {a}{\dots }}}}}}}}
a
3
{\displaystyle {\sqrt[{3}]{a}}}
a
⋅
a
⋅
a
⋅
⋯
⋅
a
=
a
2
n
−
1
2
n
{\displaystyle {\sqrt {a\cdot {\sqrt {a\cdot {\sqrt {a\cdot \dots \cdot {\sqrt {a}}}}}}}}=a^{\frac {2^{n}-1}{2^{n}}}}
Pecahan bersusun berulang
Bentuk:
a
b
+
c
a
b
+
c
a
b
+
c
…
{\displaystyle {\frac {a}{b+c{\frac {a}{b+c{\frac {a}{b+c\dots }}}}}}}
a
+
b
+
a
b
a
+
b
+
a
b
a
+
b
+
…
=
a
dimana a lebih besar dari b
{\displaystyle a+b+{\frac {ab}{a+b+{\frac {ab}{a+b+\dots }}}}=a{\text{ dimana a lebih besar dari b }}}
±
(
a
+
b
)
−
a
b
±
(
a
+
b
)
−
a
b
±
(
a
+
b
)
−
…
=
±
a
atau
±
b
{\displaystyle \pm (a+b)-{\frac {ab}{\pm (a+b)-{\frac {ab}{\pm (a+b)-\dots }}}}=\pm a{\text{ atau }}\pm b}
(ax+b)(cx+d) = acx2 +(ad+bc)x+bd
(ax+b)2 = a2 x2 +2abx+b2
(ax-b)2 = a2 x2 -2abx+b2
(ax+b)(ax-b) = a2 x2 -b2 x2
(
a
±
b
)
2
=
a
2
+
b
2
±
2
a
b
{\displaystyle ({\sqrt {a}}\pm {\sqrt {b}})^{2}=a^{2}+b^{2}\pm 2{\sqrt {ab}}}
1
a
=
1
a
⋅
a
a
=
a
a
{\displaystyle {\frac {1}{\sqrt {a}}}={\frac {1}{\sqrt {a}}}\cdot {\frac {\sqrt {a}}{\sqrt {a}}}={\frac {\sqrt {a}}{a}}}
1
a
±
b
=
1
a
±
b
⋅
a
∓
b
a
∓
b
=
a
∓
b
a
−
b
{\displaystyle {\frac {1}{{\sqrt {a}}\pm {\sqrt {b}}}}={\frac {1}{{\sqrt {a}}\pm {\sqrt {b}}}}\cdot {\frac {{\sqrt {a}}\mp {\sqrt {b}}}{{\sqrt {a}}\mp {\sqrt {b}}}}={\frac {{\sqrt {a}}\mp {\sqrt {b}}}{a-b}}}
Hukum Pascal
(
a
±
b
)
0
=
1
{\displaystyle (a\pm b)^{0}=1}
(
a
±
b
)
1
=
a
±
b
{\displaystyle (a\pm b)^{1}=a\pm b}
(
a
±
b
)
2
=
a
2
±
2
a
b
+
b
2
{\displaystyle (a\pm b)^{2}=a^{2}\pm 2ab+b^{2}}
(
a
±
b
)
3
=
a
3
±
3
a
2
b
+
3
a
b
2
±
b
3
{\displaystyle (a\pm b)^{3}=a^{3}\pm 3a^{2}b+3ab^{2}\pm b^{3}}
(
a
±
b
)
4
=
a
4
±
4
a
3
b
+
6
a
2
b
2
±
4
a
b
3
+
b
4
{\displaystyle (a\pm b)^{4}=a^{4}\pm 4a^{3}b+6a^{2}b^{2}\pm 4ab^{3}+b^{4}}
dst
Pembagian istimewa
1
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
{\displaystyle a^{2}-b^{2}=(a-b)(a+b)}
a
3
−
b
3
=
(
a
−
b
)
(
a
2
+
a
b
+
b
2
)
{\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})}
a
4
−
b
4
=
(
a
−
b
)
(
a
3
+
a
2
b
+
a
b
2
+
b
3
)
=
(
a
−
b
)
(
a
+
b
)
(
a
2
+
b
2
)
{\displaystyle a^{4}-b^{4}=(a-b)(a^{3}+a^{2}b+ab^{2}+b^{3})=(a-b)(a+b)(a^{2}+b^{2})}
a
n
−
b
n
=
(
a
−
b
)
(
a
n
−
1
+
a
n
−
2
b
+
a
n
−
3
b
2
+
⋯
+
a
2
b
n
−
3
+
a
b
n
−
2
+
b
n
−
1
)
{\displaystyle a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+\dots +a^{2}b^{n-3}+ab^{n-2}+b^{n-1})}
2
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
{\displaystyle a^{2}-b^{2}=(a+b)(a-b)}
a
4
−
b
4
=
(
a
+
b
)
(
a
3
−
a
2
b
+
a
b
2
−
b
3
)
{\displaystyle a^{4}-b^{4}=(a+b)(a^{3}-a^{2}b+ab^{2}-b^{3})}
a
6
−
b
6
=
(
a
+
b
)
(
a
5
−
a
4
b
+
a
3
b
2
−
a
2
b
3
+
a
b
4
−
b
5
)
{\displaystyle a^{6}-b^{6}=(a+b)(a^{5}-a^{4}b+a^{3}b^{2}-a^{2}b^{3}+ab^{4}-b^{5})}
a
n
−
b
n
=
(
a
+
b
)
(
a
n
−
1
−
a
n
−
2
b
+
a
n
−
3
b
2
−
⋯
−
a
2
b
n
−
3
+
a
b
n
−
2
−
b
n
−
1
)
{\displaystyle a^{n}-b^{n}=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^{2}-\dots -a^{2}b^{n-3}+ab^{n-2}-b^{n-1})}
3
a
3
+
b
3
=
(
a
+
b
)
(
a
2
−
a
b
+
b
2
)
{\displaystyle a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})}
a
5
+
b
5
=
(
a
+
b
)
(
a
4
−
a
3
b
+
a
2
b
2
−
a
b
3
+
b
4
)
{\displaystyle a^{5}+b^{5}=(a+b)(a^{4}-a^{3}b+a^{2}b^{2}-ab^{3}+b^{4})}
a
7
+
b
7
=
(
a
+
b
)
(
a
6
−
a
5
b
+
a
4
b
2
−
a
3
b
3
+
a
2
b
4
−
a
b
5
+
b
6
)
{\displaystyle a^{7}+b^{7}=(a+b)(a^{6}-a^{5}b+a^{4}b^{2}-a^{3}b^{3}+a^{2}b^{4}-ab^{5}+b^{6})}
a
n
+
b
n
=
(
a
+
b
)
(
a
n
−
1
−
a
n
−
2
b
+
a
n
−
3
b
2
−
⋯
+
a
2
b
n
−
3
−
a
b
n
−
2
+
b
n
−
1
)
{\displaystyle a^{n}+b^{n}=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^{2}-\dots +a^{2}b^{n-3}-ab^{n-2}+b^{n-1})}
Pembagian x
a
2
+
b
2
=
(
a
−
b
)
(
(
a
+
b
)
+
2
b
2
a
−
b
)
{\displaystyle a^{2}+b^{2}=(a-b)((a+b)+{\frac {2b^{2}}{a-b}})}
a
3
+
b
3
=
(
a
−
b
)
(
(
a
2
+
a
b
+
b
2
)
+
2
b
3
a
−
b
)
{\displaystyle a^{3}+b^{3}=(a-b)((a^{2}+ab+b^{2})+{\frac {2b^{3}}{a-b}})}
a
4
+
b
4
=
(
a
−
b
)
(
(
a
3
+
a
2
b
+
a
b
2
+
b
3
)
+
2
b
4
a
−
b
)
=
(
a
−
b
)
(
(
a
+
b
)
(
a
2
+
b
2
)
+
2
b
4
a
−
b
)
{\displaystyle a^{4}+b^{4}=(a-b)((a^{3}+a^{2}b+ab^{2}+b^{3})+{\frac {2b^{4}}{a-b}})=(a-b)((a+b)(a^{2}+b^{2})+{\frac {2b^{4}}{a-b}})}
a
n
+
b
n
=
(
a
−
b
)
(
(
a
n
−
1
+
a
n
−
2
b
+
a
n
−
3
b
2
+
⋯
+
a
2
b
n
−
3
+
a
b
n
−
2
+
b
n
−
1
)
+
2
b
n
a
−
b
)
{\displaystyle a^{n}+b^{n}=(a-b)((a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+\dots +a^{2}b^{n-3}+ab^{n-2}+b^{n-1})+{\frac {2b^{n}}{a-b}})}
![{\displaystyle {\begin{aligned}&1.{\sqrt[{3}]{64}}\\&2.{\sqrt[{3}]{474,552}}\\&3.{\sqrt[{3}]{48,228,544}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d49a312b1e832a787326e05a9c0f683610e3b52)
![{\displaystyle {\begin{aligned}&1.{\sqrt[{3}]{64}}=4\\&2.{\sqrt[{3}]{474,552}}=78\\&3.{\sqrt[{3}]{48,228,544}}=364\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6102f098468046ca443e4d3a573752a119ffa69)
![{\displaystyle {\sqrt[{3}]{a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/895424fc79dfd221f984d973ac95ca277bd0e60c)
