OSN Sekolah Menengah Atas

contoh soal

1. Berapa hasil dari $\sqrt{2015\cdot 2017\cdot 2023\cdot 2025+64}$?

Jawaban

$$\begin{array}{c}\text{misalkan 2020 = p}\\ \sqrt{2015\cdot 2017\cdot 2023\cdot 2025+64}& =\sqrt{(2020-5)\cdot (2020-3)\cdot (2020+3)\cdot (2020+5)+64}\\ & =\sqrt{(p-5)\cdot (p-3)\cdot (p+3)\cdot (p+5)+64}\\ & =\sqrt{(p-5)\cdot (p+5)\cdot (p-3)\cdot (p+3)+64}\\ & =\sqrt{(p^2-25)\cdot (p^2-9)+64}\\ & =\sqrt{p^4-34p^2+225+64}\\ & =\sqrt{p^4-34p^2+289}\\ & =\sqrt{(p^2-17{)}^2}\\ & =p^2-17\\ & =2020^2-17\\ & =(2000+20{)}^2-17\\ & =\mathrm{4.000.000}+80.000+400-17\\ & =\mathrm{4.080.383}\\ \end{array}$$

2. Berapa nilai x dari $\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots }}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2\dots }}}}=\frac{9}{10}$?

Jawaban

$$\begin{array}{cc}\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots }}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2\dots }}}}& =\frac{9}{10}\\ \text{misalkan untuk }\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots }}}}=p\\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots }}}}& =p\\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots }}}& =p^2\\ x^2-x-p& =p^2\\ x^2-2x+1+x-1& =p^2+p\\ (x-1{)}^2+(x-1)& =p^2+p\\ x-1& =p\\ \text{misalkan untuk }\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2\dots }}}& =q\\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2\dots }}}& =q\\ x^2\sqrt[3]{x^2\sqrt[3]{x^2\dots }}& =q^3\\ x^{2}q& =q^3\\ x^2& =q^2\\ x& =q\\ \frac{x-1}{x}& =\frac{9}{10}\\ x& =10\\ \end{array}$$

3. Berapa nilai x dari $(\frac{x}{x+10}{)}^{x+10}=\frac{1}{1024}$?

Jawaban

$$\begin{array}{cc}(\frac{x+10}{x}{)}^{-(x+10)}& =(1024{)}^{-1}\\ (\frac{x+10}{x}{)}^{x+10}& =1024\\ (\frac{x+10}{x}{)}^{x+10}& =2^{10}\\ (\frac{x+10}{x}{)}^{\frac{x+10}{10}}& =2\\ (1+\frac{10}{x}{)}^{1+\frac{x}{10}}& =2\\ (1+\frac{10}{x}{)}^{1+\frac{x}{10}}& =(\frac{1}{2}{)}^{-1}\\ (1+\frac{10}{x}{)}^{1+\frac{x}{10}}& =(1+(-\frac{1}{2}){)}^{(1+(-\frac{2}{1}))}\\ \frac{10}{x}& =-\frac{1}{2}\\ x& =-20\\ \end{array}$$

4. Berapa nilai x dari $x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4$?

Jawaban

$$\begin{array}{cc}x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}& =(\sqrt{x+\frac{1}{4}}{)}^2+2\cdot \sqrt{x+\frac{1}{4}}\cdot \frac{1}{2}+(\frac{1}{2}{)}^2\\ & =(\sqrt{x+\frac{1}{4}}+\frac{1}{2}{)}^2\\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}& =4\\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2}{)}^2}& =4\\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}& =4\\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2}{)}^2& =4\\ \sqrt{x+\frac{1}{4}}+\frac{1}{2}& =2\\ \sqrt{x+\frac{1}{4}}& =\frac{3}{2}\\ x+\frac{1}{4}& =\frac{9}{4}\\ x& =2\\ \end{array}$$

5. Berapa nilai x dari $\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0$?

Jawaban

$$\begin{array}{cc}\frac{x^3}{\sqrt{8-x^2}}+x^2-8& =0\\ \frac{x^3}{\sqrt{8-x^2}}& =8-x^2\\ x^3& =(8-x^2{)}^{\frac{3}{2}}\\ x& =(8-x^2{)}^{\frac{1}{2}}\\ x^2& =8-x^2\\ 2x^2-8& =0\\ x^2-4& =0\\ (x-2)(x+2)& =0\\ \text{membuktikan }\\ x=2\text{ maka hasilnya 0 }\\ x=-2\text{ maka hasilnya -8 }\\ \text{jadi }x=2\\ \end{array}$$

6. Berapa nilai x dari $\frac{2x-4}{3}+2x-6=\frac{3}{2x-4}+\frac{1}{2x-6}$?

Jawaban

$$\begin{array}{cc}\frac{2x-4}{3}+2x-6& =\frac{3}{2x-4}+\frac{1}{2x-6}\\ \frac{2x-4}{3}+2x-6& =\frac{1}{\frac{2x-4}{3}}+\frac{1}{2x-6}\\ \text{misalkan }\frac{2x-4}{3}=y\text{ dan }2x-6=z\\ y+z& =\frac{1}{y}+\frac{1}{z}\\ y+z& =\frac{y+z}{yz}\\ y+z-\frac{y+z}{yz}& =0\\ (y+z)(1-\frac{1}{yz})& =0\\ \ast y+z=0\\ y+z& =0\\ y& =-z\\ \frac{2x-4}{3}& =-(2x-6)\\ \frac{2(x-2)}{3}& =-2(x-3)\\ \frac{x-2}{3}& =-x+3\\ x-2& =-3x+9\\ 4x& =11\\ x& =\frac{11}{4}\\ \ast 1-\frac{1}{yz}=0\\ 1-\frac{1}{yz}& =0\\ yz-1& =0\\ yz& =1\\ (\frac{2x-4}{3})(2x-6)& =1\\ \frac{4x^2-20x+24}{3}& =1\\ 4x^2-20x+24& =3\\ 4x^2-20x+21& =0\\ (2x-7)(2x-3)& =0\\ x=\frac{7}{2}& \text{ atau }x=\frac{3}{2}\\ \end{array}$$

7. Berapa nilai x dari $\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}}=5$?

Jawaban

$$\begin{array}{cc}\sqrt[5]{\frac{x^{50}+x^{60}+x^{70}}{31}}& =5\\ \frac{x^{50}+x^{60}+x^{70}}{31}& =5^5\\ x^{50}+x^{60}+x^{70}& =5^5\cdot 31\\ x^{50}(1+x^{10}+x^{20})& =5^5\cdot 31\\ (x^{10}{)}^5(1+x^{10}+(x^{10}{)}^2)& =5^5\cdot 31\\ \text{ misalkan }{x}^{10}=y\\ y^5(1+y+y^2)& =5^5\cdot 31\\ y& =5\\ x^{10}& =5\\ x& {=}^{5}log10\\ \end{array}$$

8. Berapa nilai x dari $\sqrt{3x+5+\sqrt{4x+5}}=x$?

Jawaban

$$\begin{array}{cc}\sqrt{3x+5+\sqrt{4x+5}}& =x\\ \sqrt{4x+5+\sqrt{4x+5}-x}& =x\\ \text{misalkan }\sqrt{4x+5}=y\text{ dan }4x+5=y^2\\ \sqrt{4x+5+\sqrt{4x+5}-x}& =x\\ \sqrt{y^2+y-x}& =x\\ y^2+y& =x^2+x\\ y=x\\ 4x+5& =y^2\\ 4x+5& =x^2\\ x^2-4x-5& =0\\ (x-5)(x+1)& =0\\ x=5& \text{ atau }x=-1\text{ (TM) }\\ \end{array}$$

9. Berapa nilai x dari $\sqrt{1+\sqrt{1+x}}=\sqrt[3]{x}$?

Jawaban

$$\begin{array}{cc}\sqrt{1+\sqrt{1+x}}& =\sqrt[3]{x}\\ \sqrt[3]{x}& =n\\ x& =n^3\\ \sqrt{1+\sqrt{1+n^3}}& =n\\ 1+\sqrt{1+n^3}& =n^2\\ \sqrt{1+n^3}& =n^2-1\\ 1+n^3& =n^4-2n^2+1\\ n^4-n^3-2n^2& =0\\ n^2(n^2-n-2)& =0\\ n^2(n-2)(n+1)& =0\\ n=0,n=2\text{ atau }n=-1\\ n& =0\\ x& =0^3\\ & =0\\ n& =2\\ x& =2^3\\ & =8\\ n& =-1\\ x& =(-1{)}^3\\ & =-1\\ \text{yang paling mungkin untuk nilai x adalah }8\\ \end{array}$$

10. Berapa nilai x dari $\frac{\sqrt{{(}^{4}logx{)}^2{+}^{2}logx-4}}{16{-}^{2}logx-{(}^{4}logx{)}^2}=\frac{1}{4}$?

Jawaban

$$\begin{array}{cc}{}^{4}logx& =\frac{{}^{2}logx}{2}\\ \frac{\sqrt{(\frac{{}^{2}logx}{2}{)}^2{+}^{2}logx-4}}{16{-}^{2}logx-(\frac{{}^{2}logx}{2}{)}^2}& =\frac{1}{4}\\ \text{misalkan }y& {=}^{2}logx\\ \frac{\sqrt{(\frac{y}{2}{)}^2+y-4}}{16-y-(\frac{y}{2}{)}^2}& =\frac{1}{4}\\ \frac{\sqrt{(\frac{y}{2}{)}^2+y-4}}{12-(-4+y+(\frac{y}{2}{)}^2)}& =\frac{1}{4}\\ \text{misalkan }a& =(\frac{y}{2}{)}^2+y-4\\ \frac{\sqrt{a}}{12-a}& =\frac{1}{4},a\ge 0\\ \text{misalkan }\sqrt{a}& =b,b\ge 0\\ \frac{b}{12-b^2}& =\frac{1}{4}\\ 4b& =12-b^2\\ b^2+4b-12& =0\\ (b-2)(b+6)& =0\\ b=2& \text{ atau }b=-6(TM)\\ \sqrt{a}& =2\\ a& =4\\ (\frac{y}{2}{)}^2+y-4& =4\\ \frac{y^2}{4}+y-8& =0\\ y^2+4y-32& =0\\ (y-4)(y+8)& =0\\ y=4& \text{ atau }y=-8\\ {}^{2}logx& =4\\ x& =16\\ {}^{2}logx=-8\\ x& =\frac{1}{256}\\ \end{array}$$

11. Berapa nilai x dari $\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}$?

Jawaban

$$\begin{array}{cc}\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}& =\frac{\sqrt{1+x}}{\sqrt{x}}\\ \sqrt{x}(\sqrt{1+x}+\sqrt{x})& =(\sqrt{1+x}-\sqrt{x})\sqrt{1+x}\\ \sqrt{x(1+x)}+x& =1+x-\sqrt{x(1+x)}\\ 2\sqrt{x(1+x)}& =1\\ \sqrt{x(1+x)}& =\frac{1}{2}\\ x(1+x)& =\frac{1}{4}\\ x^2+x& =\frac{1}{4}\\ 4x^2+4x& =1\\ 4x^2+4x-1& =0\\ x& =\frac{-4\pm \sqrt{4^2-4(4)(-1)}}{2(4)}\\ & =\frac{-4\pm \sqrt{32}}{8}\\ & =\frac{-4\pm 4\sqrt{2}}{8}\\ & =\frac{-1\pm \sqrt{2}}{2}\\ \text{karena akar x harus minimal nol jadi }x=\frac{-1+\sqrt{2}}{2}\\ \end{array}$$

12. Berapa nilai x dari $\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}$?

Jawaban

$$\begin{array}{cc}\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}& =\frac{11}{19}\\ \text{misalkan }\sqrt{x+1}=y\text{ dan }x=y^2-1\\ \frac{y^2-1-y}{y^2-1+y}& =\frac{11}{19}\\ 19(y^2-y-1)& =11(y^2+y-1)\\ 19y^2-19y-19& =11y^2+11y-11\\ 8y^2-30y-8& =0\\ 4y^2-15y-4& =0\\ (4y+1)(y-4)& =0\\ y=-\frac{1}{4}\text{ (TM) atau }& y=4\\ x& =4^2-1\\ & =15\\ \end{array}$$

13. Berapa nilai x dari $\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98$?

Jawaban

$$\begin{array}{cc}\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}& =98\\ \text{misalkan }\sqrt{x^2-1}=y\\ \frac{x+y}{x-y}+\frac{x-y}{x+y}& =98\\ \frac{(x+y{)}^2+(x-y{)}^2}{(x-y)(x+y)}& =98\\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2}& =98\\ \frac{2(x^2+y^2)}{x^2-y^2}& =98\\ \frac{x^2+y^2}{x^2-y^2}& =49\\ x^2+y^2& =49(x^2-y^2)\\ x^2+y^2& =49x^2-49y^2\\ 48x^2& =50y^2\\ 24x^2& =25y^2\\ 24x^2& =25(\sqrt{x^2-1}{)}^2\\ 24x^2& =25(x^2-1)\\ 24x^2& =25x^2-25\\ x^2& =25\\ x& =\pm 5\\ \end{array}$$

14. Berapa nilai x dari $\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}$?

Jawaban

$$\begin{array}{c}\text{misalkan }\sqrt{x}=y\text{ dan }x=y^2\\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}& =\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}\\ \sqrt{y^2+y}-\sqrt{y^2-y}& =\frac{5}{4}\sqrt{\frac{y^2}{y^2+y}}\\ \sqrt{y^2+y}-\sqrt{y^2-y}& =\frac{5}{4}\frac{y}{\sqrt{y^2+y}}\\ y^2+y-\sqrt{(y^2+y)(y^2-y)}& =\frac{5}{4}y\\ y^2+y-\sqrt{y^4-y^2}& =\frac{5}{4}y\\ y^2+y-\sqrt{y^2(y^2-1)}& =\frac{5}{4}y\\ y(y+1)-y\sqrt{y^2-1}& =\frac{5}{4}y\\ y+1-\sqrt{y^2-1}& =\frac{5}{4}\\ -\sqrt{y^2-1}& =\frac{1}{4}-y\\ y^2-1& =(\frac{1}{4}-y{)}^2\\ y^2-1& =\frac{1}{16}-\frac{1}{2}y+y^2\\ -1& =\frac{1}{16}-\frac{1}{2}y\\ \frac{1}{2}y& =\frac{1}{16}+1\\ \frac{1}{2}y& =\frac{17}{16}\\ y& =\frac{17}{8}\\ x& =(\frac{17}{8}{)}^2\\ & =\frac{289}{64}\\ \end{array}$$

15. Berapa nilai x dari $\sqrt[4]{62+x}+\sqrt[4]{275-x}=7$?

Jawaban

$$\begin{array}{c}\text{ misalkan }\sqrt[4]{62+x}=a,62+x=a^4,\sqrt[4]{275-x}=b\text{ dan }275-x=b^4\\ a+b& =7\\ (a+b{)}^2& =49\\ a^2+b^2+2ab& =49\\ a^2+b^2& =49-2ab\\ a^4+b^4& =62+x+275-x\\ (a^2+b^2{)}^2-2(ab{)}^2& =337\\ (49-2ab{)}^2-2(ab{)}^2& =337\\ 2401-196ab+4(ab{)}^2-2(ab{)}^2& =337\\ 2(ab{)}^2-196ab+2064& =0\\ (ab{)}^2-98ab+1032& =0\\ (ab-12)(ab-86)& =0\\ ab=12\text{ atau }& ab=86\text{ (TM) karena hasil kali maksimum yaitu 12 }\\ ab=12\text{ dan }a+b=7\\ a+b& =7\\ b& =7-a\\ ab& =12\\ a(7-a)& =12\\ -a^2+7a& =12\\ a^2-7a+12& =0\\ (a-3)(a-4)& =0\\ a=3\text{ atau }& a=4\\ a=3,b=4\\ 62+x& =a^4\\ 62+x& =(3{)}^4\\ 62+x& =81\\ x& =19\\ a=4,b=3\\ 62+x& =a^4\\ 62+x& =(4{)}^4\\ 62+x& =256\\ x& =194\\ \end{array}$$

16. Berapa nilai x dari $\sqrt[3]{(8+x{)}^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x{)}^2}=7$?

Jawaban

$$\begin{array}{cc}\sqrt[3]{(8+x{)}^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x{)}^2}& =7\\ (\sqrt[3]{8+x}{)}^2-\sqrt[3]{8+x}\sqrt[3]{27-x}+(\sqrt[3]{27-x}{)}^2& =7\\ \text{misalkan }\sqrt[3]{8+x}=a,8+x=a^3,\sqrt[3]{27-x}=b\text{ dan }27-x=b^3\\ a^2-ab+b^2& =7\\ a^3+b^3& =8+x+27-x\\ & =35\\ a^3+b^3& =(a+b)(a^2-ab+b^2)\\ 35& =(a+b)(7)\\ a+b& =5\\ b& =5-a\\ (a+b{)}^3& =a^3+b^3+3ab(a+b)\\ 5^3& =35+3ab(5)\\ 125& =35+15ab\\ 80& =15ab\\ ab& =6\\ a(5-a)& =6\\ 5a-a^2& =6\\ a^2-5a+6& =6\\ (a-2)(a-3)& =6\\ a=2& \text{ atau }a=3\\ a=2,b=3\text{ dan }a=3,b=2\\ 8+x& =a^3\\ & =2^3\\ & =8\\ x& =0\\ 8+x& =a^3\\ & =3^3\\ & =27\\ x& =19\\ \end{array}$$

17. Berapa nilai x dari 4 . 9^x+3 . 16^x = 7 . 12^x?

Jawaban

$$\begin{array}{c}\text{kemungkinan 1}\\ 4\cdot 9^x+3\cdot 16^x& =7\cdot 12^x\\ 4\cdot (3^2{)}^x+3\cdot (4^2{)}^x& =7\cdot (4\cdot 3{)}^x\\ 4\cdot 3^{2x}+3\cdot 4^{2x}& =7\cdot 4^x\cdot 3^x\\ 4\cdot (3^x{)}^2+3\cdot (4^x{)}^2& =7\cdot 4^x\cdot 3^x\\ 4+3\cdot ((\frac{4}{3}{)}^x{)}^2& =7\cdot (\frac{4}{3}{)}^x\\ \text{misalkan }(\frac{4}{3}{)}^x=y\\ 4+3y^2& =7y\\ 3y^2-7y+4& =0\\ (3y-4)(y-1)& =0\\ y_1=\frac{4}{3}& \text{ atau }{y}_2=1\\ \ast y_1=\frac{4}{3}\\ (\frac{4}{3}{)}^x& =\frac{4}{3}\\ x& =1\\ \ast y_2=1\\ (\frac{4}{3}{)}^x& =1\\ x& =0\\ \text{kemungkinan 2}\\ 4\cdot 9^x+3\cdot 16^x& =7\cdot 12^x\\ 4\cdot (3^2{)}^x+3\cdot (4^2{)}^x& =7\cdot (4\cdot 3{)}^x\\ 4\cdot 3^{2x}+3\cdot 4^{2x}& =7\cdot 4^x\cdot 3^x\\ 4\cdot (3^x{)}^2+3\cdot (4^x{)}^2& =7\cdot 4^x\cdot 3^x\\ 4\cdot ((\frac{3}{4}{)}^x{)}^2+3& =7\cdot (\frac{3}{4}{)}^x\\ \text{misalkan }(\frac{3}{4}{)}^x=y\\ 4y^2+3& =7y\\ 4y^2-7y+3& =0\\ (4y-3)(y-1)& =0\\ y_1=\frac{3}{4}& \text{ atau }{y}_2=1\\ \ast y_1=\frac{3}{4}\\ (\frac{3}{4}{)}^x& =\frac{3}{4}\\ x& =1\\ \ast y_2=1\\ (\frac{3}{4}{)}^x& =1\\ x& =0\\ \end{array}$$

18. Berapa nilai x dari $3^x+5^x-9^x+15^x-25^x=1$?

Jawaban

$$\begin{array}{cc}{3}^x+5^x-9^x+15^x-25^x& =1\\ 3^x+5^x-(3^2{)}^x+(3\cdot 5{)}^x-(5^2{)}^x& =1\\ 3^x+5^x-(3^x{)}^2+(3^x\cdot 5^x)-(5^x{)}^2& =1\\ \text{misalkan }{3}^x=a\text{ dan }{5}^x=b\\ a+b-a^2+ab-b^2& =1\\ a^2-ab+b^2-a-b+1& =0\\ 2a^2-2ab+2b^2-2a-2b+2& =0\\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1& =0\\ (a-b{)}^2+(a-1{)}^2+(b-1{)}^2& =0\\ a-b=0;a-1=0;b-1& =0\\ a=b& =1\\ 3^x& =1\\ x& =0\\ \end{array}$$

19. Berapa nilai x dari ${}^{6}log{x}^2{+}^{6x}log\frac{6}{x}=1$?

Jawaban

$$\begin{array}{cc}{}^{6}log{x}^2{+}^{6x}log\frac{6}{x}& =1\\ \text{misalkan }6x=a\text{ maka }x=\frac{a}{6}\\ {}^{6}log{x}^2{+}^{6x}log\frac{6}{x}& =1\\ {}^{6}log(\frac{a}{6}{)}^2{+}^{6\frac{a}{6}}log\frac{6}{\frac{a}{6}}& =1\\ {}^{6}log\frac{a^2}{6^2}{+}^{a}log\frac{6^2}{a}& =1\\ {}^{6}log{a}^2{-}^{6}log{6}^2{+}^{a}log{6}^2{-}^{a}loga& =1\\ 2^{6}loga-2^{6}log6+2^{a}log6{-}^{a}loga& =1\\ 2^{6}loga-2+2\frac{1}{{}^{6}loga}-1& =1\\ 2^{6}loga+2\frac{1}{{}^{6}loga}-4& =0\\ 2^{6}lo{g}^{2}a-4^{6}loga+2& =0\\ {}^{6}lo{g}^{2}a-2^{6}loga+1& =0\\ {(}^{6}loga-1{)}^2& =0\\ {}^{6}loga& =1\\ a& =6\\ x& =\frac{a}{6}\\ & =\frac{6}{6}\\ & =1\\ \end{array}$$

20. Berapa nilai x dari (x+500)³+x=20?

Jawaban

$$\begin{array}{cc}(x+500{)}^3+x& =20\\ \text{misalkan }a=x+500\text{ maka }x=a-500\\ a^3+a-500& =20\\ a^3+a& =520\\ a(a^2+1)& =8\cdot 65\\ a(a^2+1)& =8(64+1)\\ a(a^2+1)& =8(8^2+1)\\ a& =8\\ x& =8-500\\ & =-492\\ \end{array}$$

21. Berapa nilai x dari $\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0$?

Jawaban

$$\begin{array}{cc}\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}& =0\\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}& =\frac{1}{2}\\ \frac{x^n+4^n}{x^n+16^n}& =(\frac{1}{2}{)}^n\\ \frac{x^n+4^n}{x^n+16^n}& =\frac{1}{2^n}\\ 2^n(x^n+4^n)& =x^n+16^n\\ 2^n(x^n+2^{2n})& =x^n+2^{4n}\\ 2^n\cdot x^n+2^{3n}& =x^n+2^{4n}\\ 2^n\cdot x^n-x^n& =2^{4n}-2^{3n}\\ x^n(2^n-1)& =2^{3n}(2^n-1)\\ x^n& =2^{3n}\\ x^n& =(2^3{)}^n\\ x^n& =8^n\\ x& =8\\ \end{array}$$

22. Berapa hasil dari $\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}$?

Jawaban

$$\begin{array}{c}\text{misalkan }x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}\\ x& =\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}\\ & =\frac{\sqrt{5\cdot 6}+\sqrt{5\cdot 5}+\sqrt{6\cdot 4}+\sqrt{5\cdot 4}}{\sqrt{5\cdot 4}+\sqrt{6}+\sqrt{4}}\\ & =\frac{\sqrt{5}\cdot \sqrt{6}+\sqrt{5}\cdot \sqrt{5}+\sqrt{6}\cdot \sqrt{4}+\sqrt{5}\cdot \sqrt{4}}{2\cdot \sqrt{5}+\sqrt{6}+\sqrt{4}}\\ & =\frac{\sqrt{6}\cdot \sqrt{5}+\sqrt{6}\cdot \sqrt{4}+\sqrt{5}\cdot \sqrt{5}+\sqrt{5}\cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}}\\ & =\frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}\\ & =\frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}\\ \frac{1}{x}& =\frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}\\ & =\frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}\\ & =\frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}}\\ & =\frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5}\\ & =\frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1}\\ & =\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}\\ & =\sqrt{6}-\sqrt{4}\\ & =\sqrt{6}-2\\ x& =\frac{1}{\sqrt{6}-2}\\ & =\frac{\sqrt{6}+2}{6-4}\\ & =\frac{\sqrt{6}+2}{2}\\ & =1+\frac{\sqrt{6}}{2}\\ \end{array}$$

23. Berapa hasil dari $(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}}{)}^5$?

Jawaban

$$\begin{array}{c}(\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}\\ \ast cara1\\ 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}\\ \ast cara2\\ 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{array}$$

24. Berapa hasil dari $\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots$?

Jawaban

$$\begin{array}{cc}x& =\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4}& =\frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4}& =\frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4}& =\frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots )\\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots & =\frac{1}{1-\frac{1}{4}}\\ & =\frac{4}{3}\\ \frac{3x}{4}& =\frac{1}{4}+4(\frac{4}{3})\\ & =\frac{1}{4}+\frac{16}{3}\\ & =\frac{67}{12}\\ x& =\frac{67}{9}\\ \end{array}$$

25. Berapa nilai y-x jika $\frac{1+2+3+4+\dots +106}{4+5+6+7+\dots +109}=\frac{x}{y}$?

Jawaban

$$\begin{array}{cc}\frac{1+2+3+4+\dots +106}{4+5+6+7+\dots +109}& =\frac{x}{y}\\ \frac{\frac{106\times 107}{2}}{\frac{106}{2}(4+109)}& =\frac{x}{y}\\ \frac{53\times 107}{53\times 113}& =\frac{x}{y}\\ y-x& =113-107=6\\ \end{array}$$

26. Berapa angka satuan dari hasil 17²⁰²⁴?

Jawaban

$$\begin{array}{c}\text{Perhatikan angka satuannya}\\ 17^1& =7\\ 17^2& =9\\ 17^3& =3\\ 17^4& =1\\ 17^5& =7\\ 17^6& =9\\ 17^7& =3\\ 17^8& =1\\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1}\end{array}$$

27. Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!?

Jawaban

$$\begin{array}{c}\text{Perhatikan}\\ 1!+2!+3!+4!+\dots +2024!& =1+(1x2)+(1x2x3)+(1x2x3x4)+\dots +2024!\\ & =1+2+6+24+120+720+\dots +2024!\\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka }1+2+6+24=33\text{ jadi angka satuannya adalah }3\end{array}$$

28. Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12?

Jawaban

$$\begin{array}{c}\text{Perhatikan}\\ \frac{1!+2!+3!+4!+\dots +2024!}{12}& =\frac{1+1x2+1x2x3+1x2x3x4+\dots +2024!}{12}\\ & =\frac{1+2+6+24+\dots +2024!}{12}\\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi }1+2+6=9\end{array}$$

29. 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

$$\begin{array}{c}\text{penjumlahan}\\ 1+1+1+1+\dots \text{ (sebanyak 88 buah) }+x+y& =A\\ 88+x+y& =A\\ \text{perkalian}\\ 1\times 1\times 1\times \dots \text{ (sebanyak 88 buah) }\times x\times y& =A\\ x\times y& =A\\ 88+x+y& =xy\\ xy-y& =88+x\\ y(x-1)& =88+x\\ y& =\frac{88+x}{x-1}\\ \text{uji selidiki untuk x=2}\\ y& =\frac{88+2}{2-1}\\ & =90\\ \text{buktikan}\\ 88+x+y& =xy\\ 88+2+90& =2(90)\\ 180& =180\\ \text{terbukti}\\ \text{nilai A adalah }180\\ \end{array}$$

30. Berapakah nilai x, y dan z dari $x+y-z=1,x^2+y^2-z^2=-5\text{ dan }{x}^3+y^3-z^3=-53$?

Jawaban

$$\begin{array}{cc}x+y-z& =1\\ x+y& =z+1\\ x^2+2xy+y^2& =z^2+2z+1\\ x^2+y^2-z^2& =2z+1-2xy\\ -5& =2z+1-2xy\\ 2xy& =2z+6\\ xy& =z+3\\ x^2+y^2-z^2& =-5\\ x^2+y^2& =z^2-5\\ x^3+y^3-z^3& =-53\\ (x+y)(x^2-xy+y^2)-z^3+53& =0\\ (x+y)(x^2+y^2-xy)-z^3+53& =0\\ (z+1)(z^2-5-(z+3))-z^3+53& =0\\ (z+1)(z^2-z-8)-z^3+53& =0\\ z^3-z^2-8z+z^2-z-8-z^3+53& =0\\ -9z+45& =0\\ -9z& =-45\\ z& =5\\ x+y& =5+1\\ x+y& =6\\ x& =6-y\\ xy& =5+3\\ xy& =8\\ (6-y)y& =8\\ 6y-y^2& =8\\ y^2-6y+8& =0\\ (y-4)(y-2)& =0\\ y=4\text{ atau }y=2\\ \text{jika }y=4\\ x+y& =z+1\\ x+4& =5+1\\ x& =2\\ \text{jika }y=2\\ x+y& =z+1\\ x+2& =5+1\\ x& =4\\ \end{array}$$

31. Berapakah nilai titik koordinat (x,y) dari $\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}$ dan $\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}$?

Jawaban

$$\begin{array}{cc}\sqrt{x+y}+\sqrt{x-y}& =\sqrt{\frac{432x}{13y}}\\ \sqrt{x+y}-\sqrt{x-y}& =\sqrt{\frac{52y}{3x}}\\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y})& =\sqrt{\frac{432x}{13y}}\cdot \sqrt{\frac{52y}{3x}}\\ x+y-x+y& =\sqrt{\frac{432x\cdot 52y}{13y\cdot 3x}}\\ 2y& =\sqrt{144\cdot 4}\\ 2y& =\sqrt{576}\\ 2y& =24\\ y& =12\\ \sqrt{x+12}+\sqrt{x-12}& =\sqrt{\frac{432x}{13y}}\\ \sqrt{x+12}+\sqrt{x-12}& =\sqrt{\frac{432x}{13(12)}}\\ x+12+x-12+2\cdot \sqrt{x+12}\cdot \sqrt{x-12}& =\frac{36x}{13}\\ 2x+2\sqrt{x^2-144}& =\frac{36x}{13}\\ 2(x+\sqrt{x^2-144})& =\frac{36x}{13}\\ x+\sqrt{x^2-144}& =\frac{18x}{13}\\ \sqrt{x^2-144}& =\frac{5x}{13}\\ x^2-144& =\frac{25x^2}{169}\\ \frac{144x^2}{169}-144& =0\\ \frac{x^2}{169}-1& =0\\ x^2-169& =0\\ (x-13)(x+13)& =0\\ x_1=13& \text{ atau }{x}_2=-13\text{ (TM) karena }x>y\\ \end{array}$$

jadi titik koordinat (13,12)

32. Berapakah nilai dari $x^2-7x$ jika $(x-2{)}^2+\frac{1}{(x-2{)}^2}=11$?

Jawaban

$$\begin{array}{cc}(x-2{)}^2+\frac{1}{(x-2{)}^2}& =11\\ (x-2{)}^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2{)}^2}& =11-2\\ (x-2-\frac{1}{x-2}{)}^2& =9\\ x-2-\frac{1}{x-2}& =3\\ (x-2{)}^2-1& =3(x-2)\\ x^2-4x+4-1& =3x-6\\ x^2-7x& =-9\\ \end{array}$$

33. Berapakah nilai dari $\frac{(x+y{)}^2(x+z{)}^2(x+z{)}^2}{(x^2+1)(y^2+1)(z^2+1)}$ jika xy+yz+xz=1?

Jawaban

$$\begin{array}{cc}xy+yz+xz& =1\\ x^2+xy+yz+xz& =x^2+1\\ x(x+y)+z(x+y)& =x^2+1\\ (x+y)(x+z)& =x^2+1\\ \text{dengan pola yang sama }\\ (y+x)(y+z)& =y^2+1\\ (x+z)(y+z)& =z^2+1\\ \frac{(x+y{)}^2(y+z{)}^2(x+z{)}^2}{(x^2+1)(y^2+1)(z^2+1)}& =\frac{(x+y{)}^2(y+z{)}^2(x+z{)}^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)}\\ & =\frac{(x+y{)}^2(y+z{)}^2(x+z{)}^2}{(x+y{)}^2(y+z{)}^2(x+z{)}^2}\\ & =1\\ \end{array}$$

34. Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5?

Jawaban

$$\begin{array}{cc}w+5& =w+x+y+z+5\\ x+4& =w+x+y+z+5\\ y+3& =w+x+y+z+5\\ z+2& =w+x+y+z+5\\ \text{jumlahkan keempat persamaan }\\ w+x+y+z+14& =4(w+x+y+z+5)\\ w+x+y+z+14& =4(w+x+y+z)+20\\ 3(w+x+y+z)& =-6\\ w+x+y+z& =-2\\ \end{array}$$

35. Berapakah nilai dari $\frac{x^2{y}^2+y^2{z}^2+x^2{z}^2}{x^2{y}^2{z}^2}$ jika $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$ dan x+y+z=xyz?

Jawaban

$$\begin{array}{cc}\frac{x^2{y}^2+y^2{z}^2+x^2{z}^2}{x^2{y}^2{z}^2}& =\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}\\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}{)}^2& =\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz})\\ 3^2& =\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz})\\ 9& =\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz})\\ & =\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}& =7\\ \end{array}$$

36. Berapakah nilai dari $\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}$ jika $x^2+y^2+z^2=-2(ab+bc+ac)$?

Jawaban

$$\begin{array}{cc}{x}^2+y^2+z^2& =-2(xy+yz+xz)\\ x^2+y^2+z^2+2(xy+yz+xz)& =0\\ (x+y+z{)}^2& =0\\ x+y+z& =0\\ x+y& =-z\\ x+z& =-y\\ y+z& =-x\\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}& =\frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x}\\ & =-2-(-5)-(-7)\\ & =10\\ \end{array}$$

37. Berapakah nilai dari $\frac{20xyz}{xy+yz+xz}$ jika $16^x=256^y=625^z=40$?

Jawaban

$$\begin{array}{cc}16^x=256^y=625^z& =40\\ 2^{4x}=4^{4y}=5^{4z}& =40\\ 2^{4x}& =40\\ 2& =40^{\frac{1}{4x}}\\ 4^{4y}& =40\\ 4& =40^{\frac{1}{4y}}\\ 5^{4z}& =40\\ 5& =40^{\frac{1}{4z}}\\ 2\cdot 4\cdot 5& =40^{\frac{1}{4x}}\cdot 40^{\frac{1}{4y}}\cdot 40^{\frac{1}{4z}}\\ 40& =40^{\frac{1}{4x}}\cdot 40^{\frac{1}{4y}}\cdot 40^{\frac{1}{4z}}\\ 40& =40^{\frac{1}{4x}+\frac{1}{4y}+\frac{1}{4z}}\\ 1& =\frac{1}{4x}+\frac{1}{4y}+\frac{1}{4z}\\ 4& =\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\\ \frac{20xyz}{xy+yz+xz}& =20\cdot \frac{xyz}{xy+yz+xz}\\ & =20\cdot (\frac{xy+yz+xz}{xyz}{)}^{-1}\\ & =20\cdot (\frac{1}{z}+\frac{1}{x}+\frac{1}{y}{)}^{-1}\\ & =20\cdot (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}{)}^{-1}\\ & =20\cdot (4{)}^{-1}\\ & =20\cdot \frac{1}{4}\\ & =5\\ \end{array}$$

38. Berapakah nilai dari $\frac{x^2}{x^4+3x^2+1}$ jika $6x^2+25x+6=0$?

Jawaban

$$\begin{array}{cc}6x^2+25x+6& =0\\ 6x+25+\frac{6}{x}& =0\\ 6(x+\frac{1}{x})& =-25\\ x+\frac{1}{x}& =\frac{-25}{6}\\ (c+\frac{1}{x}{)}^2& =(\frac{-25}{6}{)}^2\\ x^2+2+\frac{1}{x^2}& =\frac{625}{36}\\ x^2+\frac{1}{x^2}& =\frac{625}{36}-2\\ x^2+\frac{1}{x^2}& =\frac{553}{36}\\ \frac{x^2}{x^4+3x^2+1}& =\frac{1}{x^2+3+\frac{1}{x^2}}\\ & =\frac{1}{a^2+\frac{1}{x^2}+3}\\ & =\frac{1}{\frac{553}{36}+3}\\ & =\frac{1}{\frac{661}{36}}\\ & =\frac{36}{661}\\ \end{array}$$

39. Berapakah nilai dari $\frac{(9+4\sqrt{5}{)}^{1013}}{(38+17\sqrt{5}{)}^{675}}+6-\sqrt{5}$?

Jawaban

$$\begin{array}{cc}\frac{(9+4\sqrt{5}{)}^{1013}}{(38+17\sqrt{5}{)}^{675}}+6-\sqrt{5}& =\frac{(9+2\sqrt{20}{)}^{1013}}{((2{)}^3+3(2{)}^2(\sqrt{5})+3(2)(\sqrt{5}{)}^2+(\sqrt{5}{)}^3{)}^{675}}+6-\sqrt{5}\\ & =\frac{((2+\sqrt{5}{)}^2{)}^{1013}}{((2+\sqrt{5}{)}^3{)}^{675}}+6-\sqrt{5}\\ & =\frac{(2+\sqrt{5}{)}^{2026}}{(2+\sqrt{5}{)}^{2025}}+6-\sqrt{5}\\ & =2+\sqrt{5}+6-\sqrt{5}\\ & =8\\ \end{array}$$

40. Berapakah nilai dari $27x^3+\frac{8}{x^3}$ jika $3x+\frac{2}{x}=6$?

Jawaban

$$\begin{array}{cc}3x+\frac{2}{x}& =6\\ (3x+\frac{2}{x}{)}^3& =6^3\\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3}& =216\\ 27x^3+18(6)+\frac{8}{x^3}& =216\\ 27x^3+108+\frac{8}{x^3}& =216\\ 27x^3+\frac{8}{x^3}& =108\\ \end{array}$$

41. Berapakah nilai dari $x^6+\frac{8}{x^3}$ jika $x^3+\frac{1}{x^3}=8$?

Jawaban

$$\begin{array}{cc}{x}^3+\frac{1}{x^3}& =8\\ x^3& =8-\frac{1}{x^3}\\ x^6& =8x^3-1\\ x^6+\frac{8}{x^3}& =8x^3-1+\frac{8}{x^3}\\ & =8x^3+\frac{8}{x^3}-1\\ & =8(x^3+\frac{1}{x^3})-1\\ & =8(8)-1\\ & =63\\ \end{array}$$

42. Berapakah nilai dari $4x+\frac{25}{x}$ jika $2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}$?

Jawaban

$$\begin{array}{cc}2\sqrt{x}+\frac{5}{\sqrt{x}}& =4x-\frac{25}{x}\\ 2\sqrt{x}+\frac{5}{\sqrt{x}}& =(2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}})\\ 1& =2\sqrt{x}-\frac{5}{\sqrt{x}}\\ 1^2& =(2\sqrt{x}-\frac{5}{\sqrt{x}}{)}^2\\ 1& =4x-20+\frac{25}{x}\\ 4x+\frac{25}{x}& =21\\ \end{array}$$

43. Berapakah nilai dari $x+\frac{1}{x}$ jika $\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}$?

Jawaban

$$\begin{array}{cc}\frac{x^2-x+1}{x^2+x+1}& =\frac{5}{6}\\ \frac{x^2+1-x}{x^2+1+x}& =\frac{5}{6}\\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}& =\frac{5}{6}\\ \text{ misalkan }x+\frac{1}{x}& =y\\ \frac{y-1}{y+1}& =\frac{5}{6}\\ 6(y-1)& =5(y+1)\\ 6y-6& =5y+5\\ y& =11\\ x+\frac{1}{x}& =11\\ \end{array}$$

44. Berapakah nilai dari $x+\frac{1}{x}$ jika $\sqrt{x}+x=1$?

Jawaban

$$\begin{array}{cc}\sqrt{x}+x& =1\\ x-1& =-\sqrt{x}\\ (x-1{)}^2& =(-\sqrt{x}{)}^2\\ x^2-2x+1& =x\\ x^2-3x+1& =0\\ x-3+\frac{1}{x}& =0\\ x+\frac{1}{x}& =3\\ \end{array}$$

45. Berapakah nilai dari $x+\frac{1}{x}$ jika $\sqrt[3]{x}-\sqrt[3]{x-36}=3$?

Jawaban

$$\begin{array}{cc}\sqrt[3]{x}-\sqrt[3]{x-36}& =3\\ (\sqrt[3]{x}-\sqrt[3]{x-36}{)}^3& =3^3\\ x-(x-36)-3\sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36})& =27\\ 36-3\sqrt[3]{x(x-36)}3& =27\\ -9\sqrt[3]{x(x-36)}& =-9\\ \sqrt[3]{x(x-36)}& =1\\ x(x-36)& =1\\ x^2-36x-1& =0\\ x-36-\frac{1}{x}& =0\\ x-\frac{1}{x}& =36\\ \end{array}$$

46. Berapakah nilai dari $x+\frac{16}{x}$ jika $x-3\sqrt{x}=4$?

Jawaban

$$\begin{array}{cc}x-3\sqrt{x}& =4\\ x-4& =3\sqrt{x}\\ x^2-8x+16& =9x\\ x^2-17x+16& =0\\ x-17+\frac{16}{x}& =0\\ x+\frac{16}{x}& =17\\ \end{array}$$

47. Berapakah nilai dari $\frac{x^2}{x^4+4}$ jika $x^2-7x+2=0$?

Jawaban

$$\begin{array}{cc}{x}^2-7x+2& =0\\ x^2+2& =7x\\ x+\frac{2}{x}& =7\\ x^2+4+\frac{4}{x^2}& =49\\ x^2+\frac{4}{x^2}& =45\\ \frac{x^4+4}{x^2}& =45\\ \frac{x^2}{x^4+4}& =\frac{1}{45}\\ \end{array}$$

48. Berapakah nilai dari $x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}$ jika $x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5$?

Jawaban

$$\begin{array}{cc}{x}^{\frac{1}{4}}+x^{-\frac{1}{4}}& =5\\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}}& =25\\ x^{\frac{1}{2}}+x^{-\frac{1}{2}}& =23\\ x+2+x^{-1}& =529\\ x+x^{-1}& =527\\ x^{\frac{1}{4}}+x^{-\frac{1}{4}}& =5\\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}}& =125\\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}}& =125\\ x^{\frac{3}{4}}+x^{-\frac{3}{4}}& =110\\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}& =x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}}\\ & =527+110\\ & =637\\ \end{array}$$

49. Berapakah nilai dari $\sqrt{8x^6+x^5+x^4+5x^3+1}$ jika $\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0$?

Jawaban

$$\begin{array}{cc}\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}& =0\\ \frac{x^2+x+1}{x^5}& =0\\ x^2+x+1& =0\\ x^2+x+1& =0\\ (x-1)(x^2+x+1)& =0(x-1)\\ x^3-1& =0\\ x^3& =1\\ x& =1\\ \sqrt{8x^6+x^5+x^4+5x^3+1}& =\sqrt{(2x^3{)}^2+x^3{x}^2+x^{3}x+5x^3+1}\\ & =\sqrt{(2(1){)}^2+(1)x^2+(1)x+5(1)+1}\\ & =\sqrt{(2{)}^2+x^2+x+5+1}\\ & =\sqrt{4+x^2+x+1+5}\\ & =\sqrt{4+0+5}\\ & =\sqrt{9}\\ & =3\\ \end{array}$$

50. Berapakah nilai dari $f(1)+f(2)+f(3)+\dots +f(99)$ jika $f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}$?

Jawaban

$$\begin{array}{cc}f(x)& =\frac{1}{\sqrt{x+1}+\sqrt{x}}\\ & =\frac{\sqrt{x+1}-\sqrt{x}}{x+1-x}\\ & =\sqrt{x+1}-\sqrt{x}\\ f(1)+f(2)+f(3)+\dots +f(98)+f(99)& =\sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+\cdot +\sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99}\\ & =\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\cdot +\sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99}\\ & =\sqrt{100}-\sqrt{1}\\ & =10-1\\ & =9\\ \end{array}$$

51. Berapakah nilai dari $5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+\dots +\frac{2024}{2025})$ jika $h(x)=\frac{3}{3+9^x}$?

Jawaban

$$\begin{array}{cc}h(x)& =\frac{3}{3+9^x}\\ h(1-x)& =\frac{3}{3+9^{1-x}}\\ & =\frac{3}{3+\frac{9}{9^x}}\\ & =\frac{9^x}{3+9^x}\\ h(x)+h(1-x)& =\frac{3}{3+9^x}+\frac{9^x}{3+9^x}\\ & =\frac{3+9^x}{3+9^x}\\ & =1\\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+\dots +(1-\frac{2}{2025})+(1-\frac{1}{2025}))\\ & 5(1+1+1+\dots +1+1)\text{ sebanyak 1012 kali }\\ & 5(1012)\\ & 5060\\ \end{array}$$

52. Berapakah nilai dari $\frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72}$?

Jawaban

$$\begin{array}{cc}\frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72}& =\frac{7^{2023}{7}^2-7^{2023}+48\times 9}{7^{2023}{7}^1+7^{2023}+8\times 9}\\ & =\frac{7^{2023}(7^2-1)+48\times 9}{7^{2023}(7^1+1)+8\times 9}\\ & =\frac{7^{2023}(49-1)+48\times 9}{7^{2023}(7+1)+8\times 9}\\ & =\frac{7^{2023}\times 48+48\times 9}{7^{2023}\times 8+8\times 9}\\ & =\frac{48(7^{2023}+9)}{8(7^{2023}+9)}\\ & =\frac{48}{8}\\ & =6\\ \end{array}$$

53. Berapakah nilai dari $tan(x+\frac{\pi }{4})$ jika $\frac{1}{cosx}-tanx=\frac{4}{5}$?

Jawaban

$$\begin{array}{cc}\frac{1}{cosx}-tanx& =\frac{4}{5}\\ secx-tanx& =\frac{4}{5}\\ se{c}^{2}x-ta{n}^{2}x& =1\\ (secx+tanx)(secx-tanx)& =1\\ (secx+tanx)\frac{4}{5}& =1\\ secx+tanx& =\frac{5}{4}\\ \text{kedua persamaan dengan cara metode eliminasi }\\ 2tanx& =\frac{5}{4}-\frac{4}{5}\\ 2tanx& =\frac{9}{20}\\ tanx& =\frac{9}{40}\\ tan(x+\frac{\pi }{4})& =\frac{tanx+tan\frac{\pi }{4}}{1-tanx\cdot tan\frac{\pi }{4}}\\ & =\frac{\frac{9}{40}+1}{1-\frac{9}{40}\cdot 1}\\ & =\frac{\frac{49}{40}}{\frac{31}{40}}\\ & =\frac{49}{31}\\ \end{array}$$

54. Berapakah nilai dari $si{n}^{3}x+cs{c}^{3}x$ jika $sinx-cscx=8$?

Jawaban

$$\begin{array}{cc}\text{ Dengan menggunakan rumus: }(a-b{)}^3& =a^3-b^3-3ab(a-b)\\ (sinx-cscx{)}^3& =si{n}^{3}x-cs{c}^{3}x-3sinxcscx(sinx-cscx)\\ 8^3& =si{n}^{3}x-cs{c}^{3}x-3sinx(\frac{1}{sinx})(8)\\ 512& =si{n}^{3}x-cs{c}^{3}x-24\\ si{n}^{3}x-cs{c}^{3}x& =512+24\\ si{n}^{3}x-cs{c}^{3}x& =536\\ \end{array}$$

55. Berapakah nilai dari $(sinx+\frac{1}{cosx}{)}^2+(cosx+\frac{1}{sinx}{)}^2$ jika $\frac{1}{sinx}+\frac{1}{cosx}=10$?

Jawaban

$$\begin{array}{cc}\frac{1}{sinx}+\frac{1}{cosx}& =10\\ \frac{1}{si{n}^{2}x}+\frac{2}{sinx\cdot cosx}+\frac{1}{co{s}^{2}x}& =100\\ (sinx+\frac{1}{cosx}{)}^2+(cosx+\frac{1}{sinx}{)}^2& =si{n}^{2}x+\frac{2sinx}{cosx}+\frac{1}{co{s}^{2}x}+co{s}^{2}x+\frac{2cosx}{sinx}+\frac{1}{si{n}^{2}x}\\ & =1+\frac{1}{si{n}^{2}x}+\frac{2(si{n}^{2}x+co{s}^{2}x)}{sinx\cdot cosx}+\frac{1}{co{s}^{2}x}\\ & =1+\frac{1}{si{n}^{2}x}+\frac{2}{sinx\cdot cosx}+\frac{1}{co{s}^{2}x}\\ & =1+100\\ & =101\\ \end{array}$$

56. Berapakah nilai dari (x-1)⁶ jika $x=\frac{4cos55^{\circ }cos25^{\circ }cos10^{\circ }+sin40^{\circ }}{sin80^{\circ }}$?

Jawaban

$$\begin{array}{cc}sin80^{\circ }& =cos10^{\circ }\\ sin80^{\circ }-cos10^{\circ }& =0\\ x& =\frac{4cos55^{\circ }cos25^{\circ }cos10^{\circ }+sin40^{\circ }}{sin80^{\circ }}\\ & =\frac{4cos55^{\circ }cos25^{\circ }cos10^{\circ }+2sin20^{\circ }cos20^{\circ }}{cos10^{\circ }}\\ & =\frac{4cos55^{\circ }cos25^{\circ }cos10^{\circ }+4sin10^{\circ }cos10^{\circ }cos20^{\circ }}{cos10^{\circ }}\\ & =4cos55^{\circ }cos25^{\circ }+4sin10^{\circ }cos20^{\circ }\\ & =2(2cos55^{\circ }cos25^{\circ }+2sin10^{\circ }cos20^{\circ })\\ & =2(cos80^{\circ }+cos30^{\circ }+sin30^{\circ }+sin(-10{)}^{\circ })\\ & =2(cos80^{\circ }+cos30^{\circ }+sin30^{\circ }-sin10^{\circ })\\ & =2(cos80^{\circ }-sin10^{\circ }+cos30^{\circ }+sin30^{\circ })\\ & =2(cos80^{\circ }-sin(90^{\circ }-80^{\circ })+\frac{\sqrt{3}}{2}+\frac{1}{2})\\ & =2(cos80^{\circ }-cos80^{\circ }+\frac{\sqrt{3}}{2}+\frac{1}{2})\\ & =2(\frac{\sqrt{3}}{2}+\frac{1}{2})\\ & =\sqrt{3}+1\\ x-1& =\sqrt{3}\\ (x-1{)}^6& =(\sqrt{3}{)}^6\\ & =27\\ \end{array}$$

57. Berapakah nilai dari x jika $x=\frac{xsin20^{\circ }-x^{2}sin10^{\circ }}{2sin20^{\circ }-sin40^{\circ }}$?

Jawaban

$$\begin{array}{cc}x& =\frac{xsin20^{\circ }-x^{2}sin10^{\circ }}{2sin20^{\circ }-sin40^{\circ }}\\ 2xsin20^{\circ }-xsin40^{\circ }& =xsin20^{\circ }-x^{2}sin10^{\circ }\\ x^{2}sin10^{\circ }+xsin20^{\circ }-xsin40^{\circ }& =0\\ x(xsin10^{\circ }+sin20^{\circ }-sin40^{\circ })& =0\\ x=0& \text{ atau }xsin10^{\circ }+sin20^{\circ }-sin40^{\circ }=0\\ xsin10^{\circ }+sin20^{\circ }-sin40^{\circ }& =0\\ xsin10^{\circ }& =sin40^{\circ }-sin20^{\circ }\\ x& =\frac{sin40^{\circ }-sin20^{\circ }}{sin10^{\circ }}\\ & =\frac{2cos30^{\circ }sin10^{\circ }}{sin10^{\circ }}\\ & =2cos30^{\circ }\\ & =\frac{2\sqrt{3}}{2}\\ & =\sqrt{3}\\ \end{array}$$

58. Berapakah nilai dari $\frac{x}{y}$ jika $\frac{x^2}{x^2-16y^2}=\frac{625}{49}$?

Jawaban

$$\begin{array}{cc}\frac{x^2}{x^2-16y^2}& =\frac{625}{49}\\ \frac{x^2-16y^2}{x^2}& =\frac{49}{625}\text{ (terbalik posisinya)}\\ 1-\frac{16y^2}{x^2}& =\frac{49}{625}\\ \frac{16y^2}{x^2}& =1-\frac{49}{625}\\ (\frac{4y}{x}{)}^2& =\frac{576}{625}\\ (\frac{4y}{x}{)}^2& =(\frac{24}{25}{)}^2\\ \frac{4y}{x}& =\frac{24}{25}\\ \frac{y}{x}& =\frac{6}{25}\\ \frac{x}{y}& =\frac{25}{6}\\ \end{array}$$

59. Berapakah nilai dari $\frac{x}{y}$ jika $\frac{x}{y}+\frac{x+10y}{y+10x}=2$ serta bilangan real untuk x dan y?

Jawaban

$$\begin{array}{cc}\frac{x}{y}+\frac{x+10y}{y+10x}& =2\\ \frac{x}{y}+\frac{\frac{x}{y}+10}{1+10\frac{x}{y}}& =2\\ \text{misalkan }\frac{x}{y}=a\\ a+\frac{a+10}{1+10a}& =2\\ a(1+10a)+a+10& =2(1+10a)\\ 10a^2+a+a+10& =2+20a\\ 10a^2-18a+8& =0\\ 5a^2-9a+4& =0\\ (5a-4)(a-1)& =0\\ a=\frac{4}{5}& \text{ atau }a=1\\ \text{jadi }\frac{x}{y}=\frac{4}{5},1\\ \end{array}$$

60. Berapakah nilai dari xy jika $x^4+y^4+x^2{y}^2=15\text{ dan }{x}^2+y^2+xy=5$?

Jawaban

$$\begin{array}{cc}{x}^2+y^2+xy& =5\\ x^2+y^2& =5-xy\\ x^4+y^4+x^2{y}^2& =15\\ (x^2{)}^2+(y^2{)}^2+2x^2{y}^2-x^2{y}^2& =15\\ (x^2+y^2{)}^2-x^2{y}^2& =15\\ (5-xy{)}^2-x^2{y}^2& =15\\ 25-10xy+x^2{y}^2-x^2{y}^2& =15\\ 25-10xy& =15\\ 10xy& =10\\ xy& =1\\ \end{array}$$

61. Berapakah nilai dari x jika $4^x=63(4^3+1)(4^6+1)(4^{12}+1)+1$?

Jawaban

$$\begin{array}{cc}{4}^x& =63(4^3+1)(4^6+1)(4^{12}+1)+1\\ 4^x-1& =63(4^3+1)(4^6+1)(4^{12}+1)\\ & =63(4^3+1)(4^6+1)(4^{12}+1)\frac{4^3-1}{4^3-1}\\ & =63(4^3+1)(4^6+1)(4^{12}+1)\frac{4^3-1}{63}\\ & =(4^3+1)(4^6+1)(4^{12}+1)(4^3-1)\\ & =(4^3-1)(4^3+1)(4^6+1)(4^{12}+1)\\ & =(4^6-1)(4^6+1)(4^{12}+1)\\ & =(4^{12}-1)(4^{12}+1)\\ & =4^{24}-1\\ 4^x& =4^{24}\\ x& =24\\ \end{array}$$

62. Berapakah nilai dari $\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}$ jika $x=\sqrt{9+4\sqrt{5}}$?

Jawaban

$$\begin{array}{cc}x& =\sqrt{9+4\sqrt{5}}\\ x& =2+\sqrt{5}\\ x^2& =9+4\sqrt{5}\\ x^2-4x& =9+4\sqrt{5}-4(2+\sqrt{5})\\ x^2-4x& =1\\ x^2& =4x+1\\ x^3& =x\cdot x^2\\ & =x(4x+1)\\ & =4x^2+x\\ & =4(4x+1)+x\\ & =16x+4+x\\ & =17x+4\\ x^4& =x\cdot x^3\\ & =x(17x+4)\\ & =17x^2+4x\\ & =17(4x+1)+4x\\ & =68x+17+4x\\ & =72x+17\\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}& =\frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1}\\ & =\frac{72x+17-85x-20+8x+2+5x+3}{2}\\ & =\frac{2}{2}\\ & =1\\ \end{array}$$

63. Berapakah nilai dari $\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}$ jika 2x-1=$\sqrt{61}$?

Jawaban

$$\begin{array}{c}\text{misalkan }\frac{x^3+1}{x^5-x^4-x^3+x^2}=p\\ p& =\frac{x^3+1}{x^5-x^4-x^3+x^2}\\ & =\frac{x^3+1}{x^5-x^4-(x^3-x^2)}\\ & =\frac{x^3+1}{x^4(x-1)-x^2(x-1)}\\ & =\frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)}\\ & =\frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)}\\ & =\frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)}\\ & =\frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)}\\ & =\frac{x^2-x+1}{x^2(x-1{)}^2}\\ & =\frac{x^2-x+1}{(x(x-1){)}^2}\\ & =\frac{x(x-1)+1}{(x(x-1){)}^2}\\ 2x-1& =\sqrt{61}\\ x& =\frac{\sqrt{61}+1}{2}\\ x-1& =\frac{\sqrt{61}-1}{2}\\ x(x-1)& =(\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2})\\ & =\frac{61-1}{4}\\ & =\frac{60}{4}\\ & =15\\ p& =\frac{x(x-1)+1}{(x(x-1){)}^2}\\ & =\frac{15+1}{15^2}\\ & =\frac{16}{15^2}\\ \sqrt{p}& =\sqrt{\frac{16}{15^2}}\\ & =\frac{4}{15}\\ \end{array}$$

64. Berapakah nilai dari $(\frac{x-3}{x}{)}^{25}$ jika $x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}$?

Jawaban

$$\begin{array}{cc}x+\sqrt[5]{8}+\sqrt[5]{2}& =1+\sqrt[5]{16}+\sqrt[5]{4}\\ x+(\sqrt[5]{2}{)}^3+\sqrt[5]{2}& =1+(\sqrt[5]{2}{)}^4+(\sqrt[5]{2}{)}^2\\ x& =(\sqrt[5]{2}{)}^4-(\sqrt[5]{2}{)}^3+(\sqrt[5]{2}{)}^2-\sqrt[5]{2}+1\\ \text{misalkan }\sqrt[5]{2}=p\\ x& =p^4-p^3+p^2-p+1\\ x& =\frac{p^5+1}{p+1}\\ (\frac{x-3}{x}{)}^{25}& =(1-\frac{3}{x}{)}^{25}\\ & =(1-\frac{3}{\frac{p^5+1}{p+1}}{)}^{25}\\ & =(1-\frac{3(p+1)}{p^5+1}{)}^{25}\\ & =(1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2}{)}^5+1}{)}^{25}\\ & =(1-\frac{(3\sqrt[5]{2}+3)}{2+1}{)}^{25}\\ & =(1-\frac{(3\sqrt[5]{2}+3)}{3}{)}^{25}\\ & =(\frac{3-(3\sqrt[5]{2}+3)}{3}{)}^{25}\\ & =(\frac{3-3\sqrt[5]{2}-3)}{3}{)}^{25}\\ & =(-\sqrt[5]{2}{)}^{25}\\ & =(-2{)}^5\\ & =-32\\ \end{array}$$

65. Berapakah nilai dari $x^{50}+x^{49}+x^{48}+x^{47}+x^{46}$ jika $x^2+x+1=0$?

Jawaban

$$\begin{array}{cc}{x}^2+x+1& =0\\ x^2+x& =-1\\ \frac{x^3-1}{x-1}& =0\\ x^3& =1\\ x& =1\\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46}& =x^{48}(x^2+x+1)+x^{45}(x^2+x)\\ & =x^{48}(0)+(x^3{)}^{15}(-1)\\ & =0+(1{)}^{15}(-1)\\ & =-1\\ \end{array}$$

66. Berapakah 2²⁴ dari $8^7+8^6+8^5+8^4+8^3+8^2+8+1=A$?

Jawaban

$$\begin{array}{cc}{8}^7+8^6+8^5+8^4+8^3+8^2+8+1& =A\\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1)& =8A\\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8& =8A\\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1& =8A+1\\ 8^8+A& =8A+1\\ 8^8& =7A+1\\ (2^3{)}^8& =7A+1\\ 2^{24}& =7A+1\\ \end{array}$$

67. Berapakah nilai dari $x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1$ jika $x+\frac{1}{x}=\sqrt{3}$?

Jawaban

$$\begin{array}{cc}x+\frac{1}{x}& =\sqrt{3}\\ x^2+2+\frac{1}{x^2}& =3\\ x^2-1+\frac{1}{x^2}& =0\\ x^2(x^2-1+\frac{1}{x^2})& =x^2(0)\\ x^4-x^2+1& =0\\ (x^2+1)(x^4-x^2+1)& =(x^2+1)0\\ x^6-x^4+x^2+x^4-x^2+1& =0\\ x^6+1& =0\\ x^6& =-1\\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1& ={x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1\\ & =(-1{)}^7+(-1{)}^6+(-1{)}^5+(-1{)}^4+(-1{)}^3+(-1{)}^2-1+1\\ & =-1+1-1+1-1+1-1+1\\ & =0\\ \end{array}$$

68. Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari $\frac{c-b}{a-1}$?

Jawaban

$$\begin{array}{cc}f(x)& =a{x}^2+bx+c\\ f(2)& =a(2{)}^2+2b+c=4\\ & =4a+2b+c=4\\ f(7)& =a(7{)}^2+7b+c=49\\ & =49a+7b+c=49\\ 49a+7b+c& =49\\ 4a+2b+c& =4\\ 45a+5b& =45\text{ (f(7) dikurangi f(2)) }\\ 9a+b& =9\\ b& =-9a+9\\ 4a+2b+c& =4\\ 4a+2(-9a+9)+c& =4\\ 4a-18a+18+c& =4\\ -14a+18+c& =4\\ c& =14a-14\\ \frac{c-b}{a-1}& =\frac{14a-14-(-9a+9)}{a-1}\\ & =\frac{14(a-1)+9(a-1)}{a-1}\\ & =\frac{(14+9)(a-1)}{a-1}\\ & =23\\ \end{array}$$

69. Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²?

Jawaban

$$\begin{array}{cc}(x+y{)}^3& =x^3+y^3+3xy(x+y)\\ 11^3& =242+3xy(11)\text{ (dibagi 11)}\\ 11^2& =22+3xy\\ 121& =22+3xy\\ 99& =3xy\\ xy& =33\\ (x-y{)}^2& =x^2+y^2-2xy\\ & =((x+y{)}^2-2xy)-2xy\\ & =(x+y{)}^2-4xy\\ & =11^2-4(33)\\ & =121-132\\ & =-11\\ \end{array}$$

70. Berapa f(1)+f(-1) jika $f(\frac{ax-b}{bx-a})$=x²-5x+6?

Jawaban

$$\begin{array}{c}\text{ jika}f(1)=f(\frac{ax-b}{bx-a})\\ 1& =\frac{ax-b}{bx-a}\\ bx-a& =ax-b\\ (b-a)x& =-b+a\\ & =-(b-a)\\ & =-1\\ f(1)& =x^2-5x+6\\ & =(-1{)}^2-5(-1)+6\\ & =12\\ \text{ jika}f(-1)=f(\frac{ax-b}{bx-a})\\ -1& =\frac{ax-b}{bx-a}\\ -(bx-a)& =ax-b\\ -bx+a& =ax-b\\ (-b-a)x& =-b-a\\ & =1\\ f(-1)& =x^2-5x+6\\ & =(1{)}^2-5(1)+6\\ & =2\\ f(1)+f(-1)& =12+2\\ & =14\\ \end{array}$$

71. berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)?

Jawaban

$$\begin{array}{cc}f(x)-x& =f(x-1)\\ f(x)-f(x-1)& =x\\ x=1;f(1)-f(0)& =1\\ x=2;f(2)-f(1)& =2\\ x=3;f(3)-f(2)& =3\\ x=4;f(4)-f(3)& =4\\ \dots \\ x=200;f(200)-f(199)& =200\\ \text{ jumlahkan tersebut menjadi }\\ f(200)-f(0)& =1+2+3+4+\dots +200\\ & =\frac{200\cdot 201}{2}\\ & =20.100\\ f(200)-1& =20.100\\ & =20.101\\ \end{array}$$

72. Misalkan f(x) adalah fungsi rekursif 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

$$\begin{array}{cc}f(x)+f(15-x)& =2024\\ f(15+x)& =f(x)+2020\\ \ast cara1\\ \text{ganti x dengan 15+x }\\ f(15+x)+f(-x)& =2024\\ f(15+x)-f(x)& =2020\\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut }\\ f(x)+f(-x)& =4\\ \text{lalu dikalikan 2 masing-masing menjadi }\\ 2f(x)+2f(-x)& =8\\ \text{maka }2f(2025)+2f(-2025)& =8\\ \ast cara2\\ \text{ganti x dengan -x }\\ f(-x)+f(15+x)& =2024\\ f(15+x)-f(x)& =2020\\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut }\\ f(x)+f(-x)& =4\\ \text{lalu dikalikan 2 masing-masing menjadi }\\ 2f(x)+2f(-x)& =8\\ \text{maka }2f(2025)+2f(-2025)& =8\\ \end{array}$$

73. Misalkan f suatu fungsi rekursif yang memenuhi $2f(\frac{2002}{x})+f(x)=3x$ untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(2)!

Jawaban

$$\begin{array}{cc}2f(\frac{2002}{x})+f(x)& =3x\\ \text{ganti x dengan 2 }\\ 2f(\frac{2002}{2})+f(2)& =3(2)\\ 2f(1001)+f(2)& =6\\ \text{ganti x dengan 1001 }\\ 2f(\frac{2002}{1001})+f(1001)& =3(1001)\\ 2f(2)+f(1001)& =3003\\ 2f(2)+f(1001)& =3003\\ f(1001)& =3003-2f(2)\\ 2f(1001)+f(2)& =6\\ 2(3003-2f(2))+f(2)& =6\\ 6006-4f(2)+f(2)& =6\\ 3f(2)& =6000\\ f(2)& =2000\\ \end{array}$$

74. Misalkan f suatu fungsi rekursif yang memenuhi $f(\frac{1}{x})+\frac{1}{x}f(-x)=3x$ untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)!

Jawaban

$$\begin{array}{cc}f(\frac{1}{x})+\frac{1}{x}f(-x)& =3x\\ \text{ganti x dengan 1/3 }\\ f(3)+3f(-\frac{1}{3})& =1\\ \text{ganti x dengan -3 }\\ f(-\frac{1}{3})-\frac{1}{3}f(3)& =-9\\ \text{dikalikan 3 }\\ 3f(-\frac{1}{3})-f(3)& =-27\\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut }\\ 2f(3)& =28\\ f(3)& =14\\ \end{array}$$

75. Diketahui polinom $f(7^b-1)=7^{3b}-10$. tentukan nilai f(5)!

Jawaban

$$\begin{array}{c}\ast cara1\\ f(5)& =f(7^b-1)\\ 5& =7^b-1\\ 7^b& =6\\ f(7^b-1)& =7^{3b}-10\\ & =(7^b{)}^3-10\\ f(6-1)& =6^3-10\\ f(5)& =216-10\\ & =206\\ \ast cara2\\ \text{misalkan }{7}^b-1=a\text{ maka }{7}^b=a+1\\ f(7^b-1)& =7^{3b}-10\\ & =(7^b{)}^3-10\\ f(a)& =(a+1{)}^3-10\\ f(5)& =(5+1{)}^3-10\\ & =6^3-10\\ & =216-10\\ & =206\\ \end{array}$$

76. Diketahui polinom $f(6^b-7)=6^{3b}-2\cdot 6^{2b}-4$. tentukan nilai f(-2)!

Jawaban

$$\begin{array}{c}\ast cara1\\ f(-2)& =f(6^b-7)\\ -2& =6^b-7\\ 6^b& =5\\ f(6^b-7)& =6^{3b}-2\cdot 6^{2b}-4\\ & =(6^b{)}^3-2\cdot (6^b{)}^2-4\\ f(5-7)& =5^3-2\cdot 5^2-4\\ f(-2)& =125-50-4\\ & =71\\ \ast cara2\\ \text{misalkan }{6}^b-7=a\text{ maka }{6}^b=a+7\\ f(6^b-7)& =6^{3b}-2\cdot 6^{2b}-4\\ & =(6^b{)}^3-2\cdot (6^b{)}^2-4\\ f(a)& =(a+7{)}^3-2(a+7{)}^2-4\\ f(-2)& =(-2+7{)}^3-2(-2+7{)}^2-4\\ & =5^3-2(5{)}^2-4\\ & =125-50-4\\ & =71\\ \end{array}$$

77. Jika $f(xy)=\frac{f(x)}{y}$ dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)!

Jawaban

$$\begin{array}{cc}f(10)& =7\\ f(2\cdot 5)& =7\\ f(xy)& =\frac{f(x)}{y}\\ f(2\cdot 5)& =\frac{f(2)}{5}\\ 7& =\frac{f(2)}{5}\\ f(2)& =35\\ \end{array}$$

78. Jika $f(xy)=\frac{f(x+y)}{xy}$ dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)!

Jawaban

$$\begin{array}{cc}f(15)& =16\\ f(3\cdot 5)& =16\\ f(xy)& =\frac{f(x+y)}{xy}\\ f(3\cdot 5)& =\frac{f(3+5)}{3\cdot 5}\\ f(15)& =\frac{f(8)}{15}\\ 16& =\frac{f(8)}{15}\\ f(8)& =240\\ \end{array}$$

79. Jika $f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15$ maka tentukan nilai f(16)!

Jawaban

$$\begin{array}{cc}f(x+\frac{1}{x}+6)& =x^2+\frac{1}{x^2}+15\\ & =(x+\frac{1}{x}{)}^2-2+15\\ & =(x+\frac{1}{x}{)}^2+13\\ \text{misalkan }x+\frac{1}{x}& =p\\ f(x+\frac{1}{x}+6)& =(x+\frac{1}{x}{)}^2+13\\ f(p+6)& =p^2+13\\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 }\\ f(p+6)& =p^2+13\\ f(10+6)& =10^2+13\\ f(16)& =100+13\\ & =113\\ \end{array}$$

80. Tentukan nilai x jika $f(x)=\frac{4}{4-x}$ dan $f(x\cdot f(x){)}^{\frac{f(4x)}{f(x)}}=256$!

Jawaban

$$\begin{array}{cc}f(x)& =\frac{4}{4-x}\\ f(4x)& =\frac{4}{4-4x}\\ \frac{f(4x)}{f(x)}& =\frac{\frac{4}{4-4x}}{\frac{4}{4-x}}\\ & =\frac{4-x}{4-4x}\\ f(x\cdot f(x))& =f(x(\frac{4}{4-x}))\\ & =f(\frac{4x}{4-x})\\ & =\frac{4}{4-(\frac{4x}{4-x})}\\ & =\frac{4}{\frac{16-4x-4x}{4-x}}\\ & =\frac{4}{\frac{16-8x}{4-x}}\\ & =\frac{4(4-x)}{4(4-4x)}\\ & =\frac{4-x}{4-4x}\\ \text{misalkan }\frac{4-x}{4-4x}& =a\\ f(x\cdot f(x){)}^{\frac{f(4x)}{f(x)}}& =256\\ a^a& =256\\ a^a& =4^4\\ a& =4\\ \frac{4-x}{4-4x}& =4\\ 4-x& =16-16x\\ 15x& =12\\ x& =\frac{4}{5}\\ \end{array}$$

81. Fungsi $f(x)=\frac{kx}{2x+1}\text{dengan }x\ne -\frac{1}{2}$. Dengan f(f(x)) = x maka tentukan nilai k!

Jawaban

$$\begin{array}{cc}f(x)& =\frac{kx}{2x+1}\\ f(f(x))& =x\\ f(\frac{kx}{2x+1})& =x\\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1}& =x\\ \frac{\frac{k^{2}x}{2x+1}}{\frac{2kx+2x+1}{2x+1}}& =x\\ \frac{k^{2}x}{2kx+2x+1}& =x\\ \frac{k^2}{2kx+2x+1}& =1\\ k^2& =2kx+2x+1\\ k^2-2kx& =2x+1\\ k^2-2kx+x^2& =x^2+2x+1\\ (k-x{)}^2& =(x+1{)}^2\\ (k-x{)}^2-(x+1{)}^2& =0\\ (k-x+x+1)(k-x-(x+1))& =0\\ k=-1& \text{ atau }k=2x+1& \text{ (TM) }\\ \end{array}$$

82. Jika n = 2023²+2024² maka berapa hasil dari $\sqrt{2n-1}$?

Jawaban

$$\begin{array}{cc}n& =2023^2+2024^2\\ & =2023^2+(2023+1{)}^2\\ \text{misalkan 2023 = p}\\ n& =p^2+(p+1{)}^2\\ & =p^2+p^2+2p+1\\ & =2p^2+2p+1\\ \sqrt{2n-1}& =\sqrt{2(2p^2+2p+1)-1}\\ & =\sqrt{4p^2+4p+2-1}\\ & =\sqrt{4p^2+4p+1}\\ & =\sqrt{(2p+1{)}^2}\\ & =2p+1\\ & =2(2023)+1\\ & =4046+1\\ & =4047\\ \end{array}$$

83. Berapakah nilai xyz dan $\frac{xy+xz}{yz}$ dari x+y=14, x+z=13 dan y+z=3?

Jawaban

$$\begin{array}{c}\text{ketiga persamaan tersebut dijumlahkan menjadi }\\ 2x+2y+2z& =30\\ x+y+z& =15\\ \text{untuk mencari x, y dan z masing-masing }\\ z& =1\\ y& =2\\ x& =12\\ xyz& =24\\ \frac{xy+xz}{yz}& =\frac{1\cdot 2+1\cdot 12}{2\cdot 12}\\ & =\frac{14}{24}\\ & =\frac{7}{12}\\ \end{array}$$

84. Tentukan nilai dari x+y+z dan yz(xy-xz) merupakan bilangan bulat positif jika xy = 2, xz = 6 dan yz = 3?

Jawaban

$$\begin{array}{cc}xy\cdot yz\cdot xz& =2\cdot 3\cdot 6\\ (xyz{)}^2& =36\\ xyz& =\pm 6\\ xyz& =6\\ \frac{xyz}{xy}& =z=\frac{6}{2}=3\\ \frac{xyz}{yz}& =x=\frac{6}{3}=2\\ \frac{xyz}{xz}& =y=\frac{6}{6}=1\\ x+y+z& =6\\ yz(xy-xz)& =3(2-6)=3(-4)=-12\\ \end{array}$$

85. Tentukan nilai dari (x-z)^y jika $\frac{xy}{x+y}=\frac{1}{3}$, $\frac{yz}{y+z}=\frac{1}{4}$ dan $\frac{xz}{x+z}=\frac{1}{9}$?

Jawaban

$$\begin{array}{cc}\frac{xy}{x+y}& =\frac{1}{3}\\ \frac{x+y}{xy}& =3\text{ (terbalik posisinya)}\\ \frac{1}{y}+\frac{1}{x}& =3\\ \frac{yz}{y+z}& =\frac{1}{4}\\ \frac{y+z}{yz}& =4\text{ (terbalik posisinya)}\\ \frac{1}{z}+\frac{1}{y}& =4\\ \frac{xz}{x+z}& =\frac{1}{9}\\ \frac{x+z}{xz}& =9\text{ (terbalik posisinya)}\\ \frac{1}{z}+\frac{1}{x}& =9\\ \text{Misalkan 1/x = a, 1/y = b dan 1/z = c}\\ a+b& =3\\ b+c& =4\\ a+c& =9\\ a+b& =3\\ b+c& =4\\ a-c& =-1\\ a-c& =-1\\ a+c& =9\\ 2a& =8\\ a& =4\\ a-c& =-1\\ 4-c& =-1\\ c& =5\\ a+b& =3\\ 4+b& =3\\ b& =-1\\ \frac{1}{x}& =4\\ x& =\frac{1}{4}\\ \frac{1}{y}& =-1\\ y& =-1\\ \frac{1}{z}& =5\\ z& =\frac{1}{5}\\ (x-z{)}^y& =(\frac{1}{4}-\frac{1}{5}{)}^{-1}\\ & =(\frac{5-4}{20}{)}^{-1}\\ & =(\frac{1}{20}{)}^{-1}\\ & =20\\ \end{array}$$

86. Tentukan nilai dari a, b dan c jika $\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}$ dan a+2b+3c=28?

Jawaban

$$\begin{array}{c}\text{misalkan k untuk semua ketiga persamaan tersebut }\\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}& =k\\ a+b& =2k\\ a+c& =4k\\ b+c& =5k\\ 2a+b+c& =6k\\ 2a+5k& =6k\\ k& =2a\\ a& =\frac{k}{2}\\ b& =\frac{3k}{2}\\ c& =\frac{7k}{2}\\ a+2b+3c& =28\\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2})& =28\\ k+6k+21k& =56\\ 28k& =56\\ k& =2\\ a& =\frac{k}{2}\\ & =\frac{2}{2}=1\\ b& =\frac{3k}{2}\\ & =\frac{3(2)}{2}=3\\ c& =\frac{7k}{2}\\ & =\frac{7(2)}{2}=7\\ \end{array}$$

87. Tentukan nilai dari (b+c)^a jika $\frac{a+b+c}{2}=\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}$?

Jawaban

$$\begin{array}{cc}\frac{a+b+c}{2}& =\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}\\ a+b+c& =2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c})\\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c}& =0\\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1& =0\\ (\sqrt{a-2}-1{)}^2+(\sqrt{b-1}-1{)}^2+(\sqrt{c}-1{)}^2& =0\\ (\sqrt{a-2}-1{)}^2& =0\\ \sqrt{a-2}-1& =0\\ \sqrt{a-2}& =1\\ a-2& =1\\ a& =3\\ (\sqrt{b-1}-1{)}^2& =0\\ \sqrt{b-1}-1& =0\\ \sqrt{b-1}& =1\\ b-1& =1\\ b& =1\\ (\sqrt{c}-1{)}^2& =0\\ \sqrt{c}-1& =0\\ \sqrt{c}& =1\\ c& =1\\ (b+c{)}^a& =(2+1{)}^3\\ & =3^3\\ & =27\\ \end{array}$$

88. x dan y merupakan bilangan tak nol. Jika xy = $\frac{x}{y}$ = x-y maka berapa nilai x+y?

Jawaban

$$\begin{array}{cc}xy& =\frac{x}{y}\\ y^2& =1\\ y^2-1& =0\\ (y-1)(y+1)& =0\\ y=1& \text{ atau }y=-1\\ \frac{x}{y}& =x-y\\ x& =xy-y^2\\ x-xy& =-y^2\\ x(1-y)& =-y^2\\ x& =\frac{-y^2}{1-y}\\ \text{cek y=1 }\\ x& =\frac{-1^2}{1-1}\\ \text{tidak memenuhi syarat }\\ \text{cek y=-1 }\\ x& =\frac{-(-1{)}^2}{1-(-1)}\\ & =\frac{-1}{2}\\ x+y& =-1-\frac{1}{2}\\ & =-\frac{3}{2}\\ \end{array}$$

89. Berapa nilai x dari $(\frac{a}{b}{)}^3+(\frac{b}{a}{)}^3=2\sqrt{x}$ jika $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$?

Jawaban

$$\begin{array}{cc}\frac{1}{a}+\frac{1}{b}& =\frac{1}{a+b}\\ \frac{a+b}{ab}& =\frac{1}{a+b}\\ (a+b{)}^2& =ab\\ a^2+2ab+b^2& =ab\\ a^2+b^2& =-ab\\ \text{misalkan }\frac{a}{b}+\frac{b}{a}=n\\ \frac{a}{b}+\frac{b}{a}& =n\\ \frac{a^2+b^2}{ab}& =n\\ a^2+b^2& =nab\\ n& =-1\\ \frac{a}{b}+\frac{b}{a}& =n\\ (\frac{a}{b}{)}^3+(\frac{b}{a}{)}^3+3(\frac{a}{b}+\frac{b}{a})& =n^3\\ (\frac{a}{b}{)}^3+(\frac{b}{a}{)}^3+3n& =n^3\\ (\frac{a}{b}{)}^3+(\frac{b}{a}{)}^3& =n^3-3n\\ & =(-1{)}^3-3(-1)\\ & =2\\ 2\sqrt{x}& =2\\ \sqrt{x}& =1\\ x& =1\\ \end{array}$$

90. Berapa nilai m dari $x^2-mx-1=0$ jika $\sqrt[3]{x_1}+\sqrt[3]{x_2}=1$?

Jawaban

$$\begin{array}{cc}\sqrt[3]{x_1}& =a\\ x_1& =a^3\\ \sqrt[3]{x_2}& =b\\ x_2& =b^3\\ \sqrt[3]{x_1}+\sqrt[3]{x_2}& =1\\ a+b& =1\\ x^2-mx-1& =0\\ x_1+x_2& =m\\ x_1\cdot x_2& =-1\\ x_1+x_2& =m\\ a^3+b^3& =m\\ x_1\cdot x_2& =-1\\ a^3\cdot b^3& =-1\\ (ab{)}^2& =(-1{)}^3\\ ab& =-1\\ (a+b{)}^3& =a^3+b^3+3ab(a+b)\\ (1{)}^3& =m+3(-1)(1)\\ 1& =m-3\\ m& =4\\ \end{array}$$

91. Berapa nilai $\frac{x_1}{x_2}$ dari $a{x}^2-18x-b=0$ jika $ab=45$?

Jawaban

$$\begin{array}{cc}ab& =45\\ b& =\frac{45}{a}\\ a{x}^2-18x-b& =0\\ a{x}^2-18x-\frac{45}{a}& =0\\ a^2{x}^2-18ax-45& =0\\ (ax-3)(ax-15)& =0\\ ax-3& =0\\ x& =\frac{3}{a}\\ ax-15& =0\\ x& =\frac{15}{a}\\ \frac{x_1}{x_2}& =\frac{\frac{3}{a}}{\frac{15}{a}}\\ & =\frac{3}{15}\\ & =\frac{1}{5}\\ \frac{x_1}{x_2}& =\frac{\frac{15}{a}}{\frac{3}{a}}\\ & =\frac{15}{3}\\ & =5\\ \end{array}$$

92. Jika $\frac{u_3}{u_1+u_2}=\frac{7}{8}$ merupakan barisan aritmetika maka berapa dari $\frac{u_2+u_3}{u_1}$?

Jawaban

$$\begin{array}{cc}\frac{u_3}{u_1+u_2}& =\frac{7}{8}\\ \frac{a+2b}{a+a+b}& =\frac{7}{8}\\ \frac{a+2b}{2a+b}& =\frac{7}{8}\\ 8(a+2b)& =7(2a+b)\\ 8a+16b& =14a+7b\\ 9b& =6a\\ b& =\frac{2a}{3}\\ \frac{u_2+u_3}{u_1}& =\frac{a+b+a+2b}{a}\\ & =\frac{2a+3b}{a}\\ & =\frac{2a+3(\frac{2a}{3})}{a}\\ & =\frac{2a+2a}{a}\\ & =\frac{4a}{a}\\ & =4\\ \end{array}$$

93. Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya?

Jawaban

$$\begin{array}{cc}\frac{7p+q}{2p+q}& =\frac{17p+q}{7p+q}\\ (7p+q{)}^2& =(17p+q)(2p+q)\\ 49p^2+14pq+q^2& =34p^2+19pq+q^2\\ 15p^2& =5pq\\ 3p& =q\\ \frac{7p+q}{2p+q}& =\frac{7p+3p}{2p+3p}\\ & =\frac{10p}{5p}\\ & =2\\ \end{array}$$

94. 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

$$\begin{array}{c}\text{rataan geometris }\\ \sqrt{a\cdot b}& =b-24\\ a\cdot b& =(b-24{)}^2\\ \text{rataan aritmatik }\\ \frac{a+b}{2}& =a+15\\ a+b& =2(a+15)\\ a+b& =2a+30\\ a& =b-30\\ a\cdot b& =(b-24{)}^2\\ (b-30)b& =(b-24{)}^2\\ b^2-30b& =b^2-48b+576\\ 18b& =576\\ b& =32\\ a& =b-30\\ & =32-30\\ & =2\\ a+b& =32+2\\ & =34\\ \end{array}$$

95. Segitiga lancip ABC dengan $\frac{a^4+b^4+c^4+a^2{b}^2}{c^2(a^2+b^2)}=2$. tentukan nilai sudut C?

Jawaban

$$\begin{array}{c}\text{syarat segitiga lancip semua sudut masing-masing kurang dari }90^{\circ }\\ c^2& =a^2+b^2-2abcosC\\ cosC& =\frac{a^2+b^2-c^2}{2ab}\\ a^4+b^4+c^4+a^2{b}^2& =2c^2(a^2+b^2)\\ a^4+b^4+a^2{b}^2+c^4& =2c^2(a^2+b^2)\\ (a^2+b^2{)}^2-a^2{b}^2+c^4& =2c^2(a^2+b^2)\\ (a^2+b^2{)}^2-2c^2(a^2+b^2)+(c^2{)}^2& =a^2{b}^2\\ (a^2+b^2-c^2{)}^2& =a^2{b}^2\\ (a^2+b^2-c^2{)}^2& =(ab{)}^2\\ a^2+b^2-c^2& =\pm ab\\ cosC& =\pm \frac{ab}{2ab}\\ & =\pm \frac{1}{2}\\ & =\frac{1}{2}\text{ (karena sudut harus kurang dari }90^{\circ })\\ C& =60^{\circ }\\ \end{array}$$

96. 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

$$\begin{array}{c}\text{misalkan panjang DA dan EA }=x\text{ dan panjang AB }=y\\ \text{luas segitiga CAB }& =\frac{CA\cdot AB}{2}\\ 45& =\frac{(x+9)(x+5)}{2}\\ 90& =x^2+14x+45\\ x^2+14x& =45\\ y^2& =(x+9{)}^2+(x+5{)}^2\\ & =x^2+18x+81+x^2+10x+25\\ & =2x^2+28x+106\\ & =2(x^2+14x)+106\\ & =2(45)+106\\ & =196\\ y& =14\\ \end{array}$$

jadi panjang BC adalah 14 cm

97. 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

$$\begin{array}{c}\text{kita cari ukuran GC yaitu }\\ \frac{GC}{AD}& =\frac{CE}{DE}\\ \frac{GC}{15}& =\frac{6}{18}\\ GC& =5\\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC }\\ & =\frac{1}{2}\cdot BC\cdot CF-\frac{1}{2}\cdot GC\cdot CE\\ & =\frac{1}{2}\cdot 15\cdot 18-\frac{1}{2}\cdot 5\cdot 6\\ & =135-15\\ & =120\\ \end{array}$$

jadi luas daerah BFEG adalah 120 cm²

98. 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

$$\begin{array}{c}\\ AI& =\frac{1}{5}AF\\ & =\frac{1}{5}10\\ & =2\\ IF& =AF-AI\\ & =10-2\\ & =8\\ \text{luas trapesium AFGD }& =\frac{(AD+EF)\cdot AF}{2}\\ & =\frac{(4+6)10}{2}\\ & =50\\ \text{luas segitiga AID }& =\frac{AI\cdot AF}{2}\\ & =\frac{(2)4}{2}\\ & =4\\ \text{luas segitiga IFG }& =\frac{IF\cdot FG}{2}\\ & =\frac{(8)6}{2}\\ & =24\\ \text{luas daerah segitiga IGD }& =\text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG }\\ & =50-4-24\\ & =22\\ \end{array}$$

jadi luas daerah segitiga IGD adalah 22 cm²

99. 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-$\pi$) cm³ maka berapa luas permukaan kerucut tersebut?

Jawaban

$$\begin{array}{c}\\ \text{volume balok}\\ V_b& =x^2(12)\\ \text{volume kerucut}\\ V_b& =\frac{1}{3}\pi x^2(12)\\ & =4\pi x^2\\ V_{b-k}& =Vb-Vk\\ 100(3-\pi )& =12x^2-4\pi x^2\\ 100(3-\pi )& =4x^2(3-\pi )\\ x^2& =25\\ x& =5\\ s& =\sqrt{12^2+5^2}\\ & =\sqrt{144+25}\\ & =\sqrt{169}\\ & =13\\ \text{luas permukaan kerucut }& =\pi r(r+s)\\ & =\pi (5)(5+13)\\ & =90\pi \\ \end{array}$$

jadi luas daerah permukaan kerucut adalah 90$\pi$ cm²

100. 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

$$\begin{array}{cc}A& =3a+1\\ B& =3b+2\\ A(A+1)+3B\\ (3a+1)(3a+1+1)+3(3b+2)\\ (3a+1)(3a+2)+9b+6\\ 9a^2+9a+2+9b+6\\ 9a^2+9a+9b+8\\ 9(a^2+a+b)+8\\ \text{sisa pembagiannya adalah }8\\ \end{array}$$

101. 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

$$\begin{array}{cc}A& =9a+7\\ B& =9b+8\\ A(A-5)+9B\\ (9a+7)(9a+7-5)+9(9b+8)\\ (9a+7)(9a+2)+81b+72\\ 81a^2+81a+14+81b+72\\ 81a^2+81a+81b+86\\ 81a^2+81a+81b+81+5\\ 81(a^2+a+b+1)+5\\ \text{sisa pembagiannya adalah }5\\ \end{array}$$

102. Jika $\left[\begin{array}{cc}3& 7\\ -1& -2\end{array}\right]$ maka berapa hasil dari A²¹+A²⁵+A⁴⁶?

Jawaban

$$\begin{array}{cc}{A}^2& =A\cdot A\\ & =\left[\begin{array}{cc}3& 7\\ -1& -2\end{array}\right]\cdot \left[\begin{array}{cc}3& 7\\ -1& -2\end{array}\right]=\left[\begin{array}{cc}2& 7\\ -1& -3\end{array}\right]\\ A^3& =A^2\cdot A\\ & =\left[\begin{array}{cc}2& 7\\ -1& -3\end{array}\right]\cdot \left[\begin{array}{cc}3& 7\\ -1& -2\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\\ & =-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ & =-I\\ A^{21}+A^{25}+A^{46}& =A^{21}\cdot (I+A^4+A^{25})\\ & =A^{21}\cdot (I+A^3\cdot A+A^{24}\cdot A)\\ & =(A^3{)}^7\cdot (I+A^3\cdot A+(A^3{)}^8\cdot A)\\ & =(-I{)}^7\cdot (I-I\cdot A+(-I{)}^8\cdot A)\\ & =-I\cdot (I-A+A)\\ & =-I\cdot I\\ & =-I\\ & =-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ & =\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\\ \end{array}$$

103. 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.

104. 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.

105. Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli!

cara 1

1. 8¹ = 1
2. 8² = 1 (8²=8¹x8¹ sama dengan 1x1)
3. 8³ = 1 (8³=8¹x8² sama dengan 1x1)
4. 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2)
5. 8⁵ = 1
6. 8ⁿ = 1 (semua n untuk bilangan asli)

Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli

cara 2

1. 8ⁿ = b mod 7

1. 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil)
2. (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya)
3. 8ⁿ = 1ⁿ mod 7
4. 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1)

Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli

106. Berapa hasil sisa dari 17⁹⁹ dibagi 5?

cara 1

1. 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0

1. 7¹ = 7 (sisa 1)
2. 7² = 49 (sisa 2)
3. 7³ = 343 (sisa 3)
4. 7⁴ = 2,401 (sisa 0)
5. 7⁵ = 16,807
6. 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

107. 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

108. Berapa hasil sisa dari 41²⁰²⁴ dibagi 33?

Jawaban

$$\begin{array}{cc}41^{2024}& =41^{2024}\text{ mod }33\\ & =(33\times 3+2{)}^{2024}\text{ mod }33\\ & =2^{2024}\text{ mod }33\\ & =2^{2020}{2}^4\text{ mod }33\\ & =(2^5{)}^{404}{2}^4\text{ mod }33\\ & =(33-1{)}^{404}{2}^4\text{ mod }33\\ & =(-1{)}^{404}{2}^4\text{ mod }33\\ & =2^4\text{ mod }33\\ & =16\text{ mod }33\\ \text{Jadi hasil sisa adalah }16\\ \end{array}$$

109. Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹?

Jawaban

$$\begin{array}{cc}99^{99}& =(3^2\times 11{)}^{99}\\ & =3^{198}\times 11^{99}\\ 243^n& =(3^5{)}^n\\ & =3^{5n}\\ \text{agar bisa membagi, maka}\\ 5n& =198\\ n& =39.6\\ \text{jadi bilangan n terbesar adalah }39\\ \end{array}$$

110. Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸?</lu>

Jawaban

$$\begin{array}{cc}88^{88}& =(8\times 11{)}^{88}\\ & =8^{88}\times 11^{88}\\ & =8^{87}\times 8\times 11^{88}\\ & =(8^3{)}^{29}\times 8\times 11^{88}\\ & =512^{29}\times 8\times 11^{88}\\ 512^n& =512^{29}\\ \text{jadi bilangan n terbesar adalah }29\\ \end{array}$$

111. 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

112. 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.

113. 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.