Lanjutan (Fungsi Eksponen Dan Logaritma)

C. Bilangan Bentuk Akar dan Pangkat Pecahan

\begin{aligned}&\textrm{Jika}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{bilangan real dan}\: \: m\: \: \textrm{bilangan bulat positif, maka}\\ &\qquad\qquad\quad \LARGE\boxed{\textbf{p}^{\textbf{n}}=\textbf{q} \quad\Leftrightarrow\quad \sqrt[\textbf{n}]{\textbf{q}}=\textbf{p}}\\ &\textrm{dengan}\\ &\quad \sqrt[n]{q}\: \: \textrm{disebut sebagai akar (radikal)}\\ &\qquad q\: \: \textrm{disebut radikan (bilangan yang ditarik akarnya)}\\ &\qquad n\: \: \textrm{adalah indeks (pangakt dari akar)} \end{aligned}.

Selanjutnya perhatikanlah tabel berikut

\begin{array}{|c|l|l|c|}\hline \textbf{No}&\textbf{\: \: \qquad Bentuk Akar}&\textbf{\: \: \: \: Pangkat pecahan}\\\hline 1&\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}&a^{n}=b\quad \Rightarrow \quad a=b^{^{^{\frac{1}{n}}}}=\sqrt[n]{b}\\\hline 2&\sqrt{\displaystyle \frac{a}{b}}=\displaystyle \frac{\sqrt{a}}{\sqrt{b}}&a> 0\quad \Rightarrow \quad \sqrt[n]{a}\geq 0\\\hline 3&a\sqrt{c}\pm b\sqrt{c}=\left ( a\pm b \right )\sqrt{c}&a< 0\begin{cases} n & \text{ genap } \Rightarrow \sqrt[n]{a}< 0\\ n & \text{ ganjil } \Rightarrow \sqrt[n]{a}\quad \textbf{bukan riil} \end{cases}\\\hline 4&\sqrt{\left ( a+b \right )\pm 2\sqrt{ab}}=\left (\sqrt{a} \pm \sqrt{b}\right )&a^{^{^{^{\frac{\begin{matrix} \\ m \end{matrix}}{\begin{matrix} n\\ \end{matrix}}}}}}=\sqrt[n]{a^{m}}\\\hline 5&\displaystyle \frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a}{b}\sqrt{b}&\\\cline{1-2} 6&\displaystyle \frac{c}{a\pm \sqrt{b}}=\frac{c}{a\pm \sqrt{b}}\times \frac{a\mp \sqrt{b}}{a\mp \sqrt{b}}=\frac{c\left ( a\mp \sqrt{b} \right )}{a^{2}- b}&\textrm{Perlu diketahui juga}\\\cline{1-2} 7&\begin{aligned}\displaystyle \frac{c}{\sqrt{a}\pm \sqrt{b}}&=\frac{c}{\sqrt{a}\pm \sqrt{b}}\times \frac{\sqrt{a}\mp \sqrt{b}}{\sqrt{a}\mp \sqrt{b}}\\ &=\frac{c\left ( \sqrt{a}\mp \sqrt{b} \right )}{a- b} \end{aligned}&a^{n}=b^{m}\Leftrightarrow a=\displaystyle b^{^{^{\frac{m}{n}}}}\Leftrightarrow a^{^{^{\frac{n}{m}}}}=b\\\cline{1-2} 8&\displaystyle \frac{a}{\sqrt[n]{b^{m}}}=\frac{a}{\sqrt[n]{b^{m}}}\times \frac{\sqrt[n]{b^{n-m}}}{\sqrt[n]{b^{n-m}}}=\frac{a}{b}\sqrt[n]{b^{n-m}}&\\\cline{1-2} 9&\begin{aligned}\displaystyle \frac{c}{\sqrt[3]{a}\pm \sqrt[3]{b}}=&\frac{c}{\sqrt[3]{a}\pm \sqrt[3]{b}}\\ &\times \frac{\sqrt[3]{a^{2}}\mp \sqrt[3]{ab}+\sqrt[3]{b^{2}}}{\sqrt[3]{a^{2}}\mp \sqrt[3]{ab}+\sqrt[3]{b^{2}}} \end{aligned}&\\\hline \end{array}.

D. Grafik Fungsi Eksponen

\LARGE{\fbox{\LARGE{\fbox{CONTOH SOAL}}}}.

\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai dari bilangan-bilangan berikut ini}! \end{array}\\ \begin{array}{llllllll}\\ .\quad\quad &a.&27^{\frac{1}{3}}&k.&\left ( \displaystyle \frac{2^{3}.3^{-2}}{2^{-5}.3} \right )^{\displaystyle \frac{1}{2}}&u.&\displaystyle \frac{\left ( a^{2}.b^{-1} \right )^{\frac{1}{2}}\sqrt{a^{6}.b^{\frac{5}{3}}.c^{-2}}}{\left ( a^{3}.b^{-5}.c^{-3} \right )^{\frac{1}{3}}}\\ &b.&32^{^{\frac{2}{5}}}&l.&\displaystyle \frac{\sqrt{2}.\sqrt[3]{8}}{(2)^{\frac{1}{3}}.\sqrt[4]{16^{2}}}&v.&\displaystyle \frac{x^{2}.y^{7}}{x^{3}.y^{5}}\\ &c.&\left ( \displaystyle \frac{9}{16} \right )^{\displaystyle \frac{3}{2}}&m.&\left ( \displaystyle \frac{\sqrt{2}.2\sqrt{6}}{\sqrt{3}.\sqrt[3]{9}} \right )^{\displaystyle \frac{1}{2}}&w.&\left ( \displaystyle \frac{2x^{3}}{y^{2}}:\frac{4x^{6}}{4y^{5}} \right ).\displaystyle \frac{3x^{2}-2y}{3y}\\ &d.&\left ( \displaystyle \frac{1}{125} \right )^{-\frac{1}{3}}&n.&\displaystyle \frac{2.3^{-\frac{1}{2}}-2+3.2^{-\frac{1}{2}}}{2.3^{-\frac{1}{2}}-3.2^{-\frac{1}{2}}}&x.&\left (\sqrt[3]{x^{2}.yz^{3}} \right ).x^{-1}.y^{-2}\\ &e.&\left ( \displaystyle \frac{2}{3} \right )^{\displaystyle \frac{2}{3}}.\left ( \displaystyle \frac{3}{2} \right )^{-\displaystyle \frac{1}{3}}&o.&-3\sqrt{6}+4\sqrt{3}-2\sqrt{81}&y.&\displaystyle \frac{\sqrt{x}.\sqrt{x^{2}y^{3}}.\sqrt{xy^{2}}}{\sqrt[4]{x}.\sqrt[3]{y}}\\ &f.&\left ( \displaystyle \frac{1}{5^{3}} \right )^{-1}.\left ( \displaystyle \frac{1}{5^{2}} \right )^{2}&p.&\sqrt{250}-\sqrt{50}+15\sqrt{2}&z.&\displaystyle \frac{\left ( x^{2} \right )^{3}}{x^{4}}:\left ( \frac{x^{3}}{\left ( x^{3} \right )^{2}} \right )^{-2} \end{array}

\begin{array}{llllllll}\\ .\quad\quad&g.&\displaystyle \frac{\sqrt{3}.\sqrt{15}}{\sqrt{5}}\quad\qquad \: \: &q.&\sqrt[2]{75}-\sqrt[4]{27}+\sqrt[3]{128}\\ &h.&\displaystyle \frac{2\sqrt{3}.\sqrt{24}}{\sqrt{7}.3\sqrt{14}}&r.&\displaystyle \frac{5\sqrt{5}+2\sqrt{5}}{5-3\sqrt{5}}\\ &i&\displaystyle \frac{\sqrt{5}.\sqrt[2]{2^{3}}}{2.\sqrt[3]{3}}&s.&\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots }}}}}}\\ &j.&\displaystyle \frac{\sqrt{3}.\sqrt[3]{2}}{\left ( \frac{4}{9} \right )^{3}.\left ( \frac{2}{3} \right )^{-2}}&t.&\left ( 4^{\frac{1}{2}} \right )^{\frac{1}{2}}\left ( 2^{-2} \right )^{-2}.\sqrt[3]{0,125}.\left ( 0,25 \right ).\displaystyle \frac{1}{2}\sqrt{\frac{1}{2}} \end{array}.

. \quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|l|l|}\hline \begin{aligned}a.\quad 27^{\frac{1}{3}}&=\left ( 3^{3} \right )^{\frac{1}{3}}\\ &=3\\ &\\ & \end{aligned}&\begin{aligned}b.\quad 32^{\frac{2}{5}}&=\left ( 2^{5} \right )^{\frac{2}{5}}\\ &=2^{2}=4\\ &\\ & \end{aligned}&\begin{aligned}c.\quad \left ( \displaystyle \frac{9}{16} \right )^{\frac{3}{2}}&=\left (\left ( \displaystyle \frac{3}{4} \right )^{2} \right )^{\frac{3}{2}}\\ &=\left ( \displaystyle \frac{3}{4} \right )^{3}=\frac{27}{64} \end{aligned}&\begin{aligned}d.\quad \left ( \displaystyle \frac{1}{125} \right )^{-\frac{1}{3}}&=\left ( 5^{-3} \right )^{-\frac{1}{3}}\\ &=5\\ & \end{aligned}\\\hline \begin{aligned}e.\quad &\left ( \frac{2}{3} \right )^{\frac{2}{3}}.\left ( \frac{3}{2} \right )^{-\frac{1}{3}}\\ &=\left ( \frac{2}{3} \right )^{\frac{2}{3}}.\left ( \frac{2}{3} \right )^{\frac{1}{3}}\\ &=\left ( \frac{2}{3} \right )^{\frac{2}{3}+\frac{1}{3}}\\ &=\frac{2}{3} \end{aligned}&\begin{aligned}h.\quad &\displaystyle \frac{2\sqrt{3}.\sqrt{24}}{\sqrt{7}.3\sqrt{14}}\\ &=\displaystyle \frac{2\sqrt{3}.\sqrt{4}.\sqrt{2}.\sqrt{3}}{\sqrt{7}.3.\sqrt{2}.\sqrt{7}}\\ &=\frac{4.\sqrt{2}.3}{7.\sqrt{2}.3}\\ &=\frac{4}{7} \end{aligned}&\begin{aligned}u.\quad &\displaystyle \frac{\left ( a^{2}b^{-1} \right )^{\frac{1}{2}}.\sqrt{a^{6}.b^{\frac{5}{3}}.c^{-2}}}{\left ( a^{3} \right )^{\frac{1}{3}}}\\ &=\displaystyle \frac{a.b^{-\frac{1}{2}}.a^{\frac{6}{2}}.b^{\frac{1}{2}.\frac{5}{3}}.c^{-\frac{2}{2}}}{a^{\frac{3}{3}}.b^{-\frac{5}{3}}.c^{-\frac{3}{3}}}\\ &=a^{3}.b^{-\frac{1}{2}+\frac{5}{6}+\frac{5}{3}}\\ &=a^{3}.b^{\frac{-3+5+10}{6}}=a^{3}.b^{2} \end{aligned}&\begin{aligned}&\textrm{Untuk Soal yang belum}\\ &\textrm{dibahas silahkan}\\ &\textrm{dikerjakan sendiri}\\ &\textrm{sebagai latihan} \end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 2.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad 3^{2x-1}=1\\ &\textrm{b}.\quad 4^{x^{2}+3x-10}=1\\ &\textrm{c}.\quad 5^{3x^{2}+2x-1}=1\\ &\textrm{d}.\quad (9)^{2x+\frac{1}{2}}.\left ( \displaystyle \frac{1}{27} \right )^{3x^{2}+2}=1\\ &\textrm{e}.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-3}.(8)^{3x+1}=1\\ &\textrm{f}.\quad \displaystyle \frac{\sqrt[3]{\left ( 0,0008 \right )^{7-2x}}}{\left ( 0,2 \right )^{-4x+5}}=1\: \: ...(\textbf{SPMB 2005})\end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|l|}\hline \begin{aligned}a.\quad 3^{2x-1}&=1\\ 3^{2x-1}&=3^{0}\\ 2x-1&=0\\ 2x&=1\\ x&=\displaystyle \frac{1}{2}\\ \textrm{HP}=&\left \{ \displaystyle \frac{1}{2} \right \}\\ &\\ &\\ & \end{aligned}&\begin{aligned}e.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-3}.(8)^{3x+1}&=1\\ (2^{3-2x}).(2^{3})^{3x+1}&=2^{0}\\ 2^{3-2x+9x+3}&=2^{0}\\ 7x+6&=0\\ 7x&=-6\\ x&=-\displaystyle \frac{6}{7}\\ \textrm{HP}=\left \{ -\displaystyle \frac{6}{7} \right \}&\\ & \end{aligned}&\begin{aligned}f.\quad \displaystyle \frac{\sqrt[3]{\left ( 0,0008 \right )^{7-2x}}}{\left ( 0,2 \right )^{-4x+5}}&=1\: \: ...(\textbf{SPMB 2005})\\ \displaystyle \frac{((0,2)^{3})^{\frac{7-2x}{3}}}{(0,2)^{-4x+5}}&=(0,2)^{0}\\ (0,2)^{7-2x-(-4x+5)}&=(0,2)^{0}\\ 7-2x+4x-5&=0\\ 2x+2&=0\\ 2&=-2\\ x&=-1\\ \textrm{HP}=\left \{ -1 \right \}& \end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 3.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad 3^{2x-5}=3^{5}\\ &\textrm{b}.\quad 2^{2x+3}.\left (\displaystyle \frac{1}{8} \right )^{x-3}=\displaystyle \frac{1}{64}\\ &\textrm{c}.\quad (2)^{5x^{2}-3x}.\left ( \displaystyle \frac{1}{32} \right )^{5}=32^{5}\\ &\textrm{d}.\quad \left ( \displaystyle \frac{1}{9} \right )^{-x^{2}}.\left ( 3^{2} \right )^{3x-3}=\displaystyle 9^{3^{2}}\\ &\textrm{e}.\quad (4)^{-2x^{2}+3x}.\left ( \displaystyle \frac{1}{4} \right )^{3}=2^{-2}.\left ( \displaystyle \frac{1}{8} \right )^{-2}\\ &\textrm{f}.\quad \displaystyle \frac{27}{3^{2x-1}}=81^{-0,125}\: \: ...(\textbf{SPMB 2004})\end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}a.\quad 3^{2x-5}&=3^{5}\\ a^{f(x)}&=a^{p}\\ f(x)&=p\\ 2x-5&=5\\ 2x&=10\\ x&=5\\ \textrm{HP}=&\left \{ 5 \right \}\\ &\\ &\\ &\\ &\\ &\end{aligned}&\begin{aligned}f.\quad \displaystyle \frac{27}{3^{2x-1}}&=81^{-0,125}\cdots (\textbf{SPMB 2004})\\ \displaystyle \frac{3^{3}}{3^{2x-1}}&=(3^{4})^{-\frac{1}{8}}\\ 3^{3-(2x-1)}&=3^{-\frac{4}{8}}\\ 3-(2x-1)&=-\frac{4}{8}\\ 4-2x&=-\frac{1}{2}\\ 8-4x&=-1\\ -4x&=-1-8\\ x&=\frac{9}{4}\\ \textrm{HP}=&\left \{ \frac{9}{4} \right \}\end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 4.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad \sqrt{3^{2x+1}}=9^{x-2}\\ &\textrm{b}.\quad \left ( \displaystyle \frac{1}{3} \right )^{2x-3}.3^{x+5}=\left ( \displaystyle \frac{1}{27} \right )^{2x-10}\\ &\textrm{c}.\quad (125)^{(x^{2}-3x-4)}=(5)^{(x^{2}-2x-3)}\\ &\textrm{d}.\quad (2^{2})^{\displaystyle \sqrt{x^{3}+2x^{2}-3x-6}}=\left ( \displaystyle \frac{1}{2} \right )^{-\displaystyle \sqrt{4x^{2}+4x-8}}\\ &\textrm{e}.\quad (4)^{(x^{2}+2x-1)}.\left ( \displaystyle \frac{1}{8} \right )^{3x-4}=\left ( \displaystyle \frac{1}{2} \right )^{2x}\\ &\textrm{f}.\quad (\sqrt{2})^{4x^{2}-8x+12}.\left ( \sqrt[3]{8} \right )^{3x+5}=\left ( \displaystyle \frac{1}{4} \right )^{4x-3}.(2)^{6x+5}\\ &\textrm{g}.\quad 9^{x^{2}-3x+1}+9^{-3x+x^{2}}=20-10\left ( 3^{x^{2}-3x} \right )\: \: ...(\textbf{SPMB 2004})\end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|l|}\hline \begin{aligned}a.\quad \sqrt{3^{2x+1}}&=9^{x-2}\\ a^{f(x)}&=a^{g(x)}\\ f(x)&=g(x)\\ \displaystyle (3)^{\frac{2x+1}{2}}&=(3^{2})^{x-2}\\ \displaystyle \frac{2x+1}{2}&=2(x-2)\\ 2x+1&=4x-8\\ -2x&=-9\\ x&=\displaystyle \frac{9}{2}\\ \textrm{HP}=&\left \{ \frac{9}{2} \right \} \end{aligned}&\begin{aligned}b.\quad \left ( \displaystyle \frac{1}{3} \right )^{2x-3}.3^{x+5}&=\left ( \displaystyle \frac{1}{27} \right )^{2x-10}\\ (3^{-1})^{(2x-3)}.3^{x+5}&=(3^{-3})^{(2x-10)}\\ 3^{(-2x+3)+(x+5)}&=3^{10-2x}\\ -2x+3+x+5&=10-2x\\ x&=10-8\\ x&=2\\ \textrm{HP}=\left \{ 2 \right \}&\\ &\\ &\\ & \end{aligned}&\begin{aligned}c.\quad (125)^{x^{2}-3x-4}&=(5)^{x^{2}-2x-3}\\ (5^{3})^{x^{2}-3x-4}&=(5)^{x^{2}-2x-3}\\ 3x^{2}-9x-12&=x^{2}-2x-3\\ 2x^{2}-7x-9&=0\\ (x+1)(2x-9)&=0\\ x=-1\: \: \textrm{V}\: \: x=\frac{9}{2}\\ \textrm{HP}=&\left \{ -1,\frac{9}{2} \right \}\\ &\\ &\\ & \end{aligned}\\\hline \multicolumn{3}{|l|}{\begin{aligned}g.\qquad\quad\quad\quad 9^{x^{2}-3x+1}+9^{-3x+x^{2}}&=20-10\left ( 3^{x^{2}-3x} \right )\\ \textrm{misalkan}\: \: 3^{x^{2}-3x}&=p\\ 9^{x^{2}-3x}.9+9^{x^{2}-3x}&=20-10\left ( 3^{x^{2}-3x} \right )\\ 9.\left ( 3^{2.(x^{2}-3x)} \right )+\left ( 3^{2.(x^{2}-3x)} \right )&=20-10\left ( 3^{x^{2}-3x} \right )\\ 10.\left ( 3^{x^{2}-3x} \right )^{2}&=20-10\left ( 3^{x^{2}-3x} \right )\\ 10p^{2}&=20-10p\\ p^{2}&=2-p\\ p^{2}+p-2&=0\\ (p+2)(p-1)&=0\\ p=-2\: (\textrm{tdk mungkin})\: \: \textrm{V}\: \: p=1&\\ \textrm{sehingga}\quad p=3^{x^{2}-3x}&=1\\ 3^{x^{2}-3x}&=3^{0}\\ x^{2}-3x&=0\\ x(x-3)&=0\\ x=1\: \: \textrm{V}\: \: x&=3\\ \textrm{HP}=&\left \{ 1,3 \right \} \end{aligned}}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 5.&\textrm{Sederhanakanlah dengan merasionalkan penyebut-penyebutnya}\\ &a.\quad \displaystyle \frac{15}{\sqrt{5}}\qquad b.\quad \displaystyle \frac{9}{2\sqrt{3}}\, \qquad c.\quad \displaystyle \sqrt{\frac{7}{2}}\, \quad\qquad d.\quad \displaystyle \frac{3\sqrt{5}}{\sqrt{72}}\\ &e.\quad \displaystyle \frac{\sqrt{50}}{\sqrt{125}}\quad f.\quad \displaystyle \frac{3}{2-\sqrt{3}}\quad g.\quad \displaystyle \frac{6}{3+2\sqrt{3}}\quad h.\quad \displaystyle \frac{3-\sqrt{2}}{3+\sqrt{2}}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a.&\begin{aligned}\displaystyle \frac{15}{\sqrt{5}}&=\displaystyle \frac{15}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}\\ &=\displaystyle \frac{15}{5}\sqrt{5}\\ &=3\sqrt{5} \end{aligned}&b.&\begin{aligned}\displaystyle \frac{9}{2\sqrt{3}}&=\displaystyle \frac{9}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ &=\displaystyle \frac{9}{2\times 3}\sqrt{3}\\ &=\displaystyle \frac{3}{2}\sqrt{3} \end{aligned}&c.&\begin{aligned}\displaystyle \sqrt{\frac{7}{2}}&=\displaystyle \sqrt{\frac{7}{2}\times \frac{2}{2}}\\ &=\displaystyle \sqrt{\frac{14}{4}}\\ &=\displaystyle \frac{1}{2}\sqrt{14} \end{aligned}&d.&\begin{aligned}\displaystyle \frac{3\sqrt{5}}{\sqrt{72}}&=\displaystyle \frac{3\sqrt{5}}{\sqrt{72}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ &=\displaystyle \frac{3\sqrt{5}.\sqrt{2}}{\sqrt{144}}\\ &=\displaystyle \frac{3\sqrt{10}}{12}\\ &=\displaystyle \frac{1}{4}\sqrt{10} \end{aligned} \\\hline \end{array} \end{array}..

.\quad\: \, \begin{aligned}&\begin{array}{|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline e.&\begin{aligned}\displaystyle \frac{50}{125}&=\displaystyle \frac{50}{125}\times \frac{5}{5}\\&=\displaystyle \frac{\sqrt{250}}{\sqrt{625}}\\ &=\displaystyle \frac{5\sqrt{10}}{25}\\ &=\displaystyle \frac{1}{5}\sqrt{10} \end{aligned}&f.&\begin{aligned}\displaystyle \frac{3}{2-\sqrt{3}}&=\displaystyle \frac{3}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}\\ &=\displaystyle \frac{3\times \left ( 2+\sqrt{3} \right )}{2^{2}-\left (\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{6+3\sqrt{3}}{4-3}\\ &=6+3\sqrt{3} \end{aligned}&g.&\begin{aligned}\displaystyle \frac{6}{3+2\sqrt{3}}&=\displaystyle \frac{6}{3+2\sqrt{3}}\times \frac{3-2\sqrt{3}}{3-2\sqrt{3}}\\ &=\displaystyle \frac{6\times \left ( 3-2\sqrt{3} \right )}{3^{2}-\left (2\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{18-12\sqrt{3}}{9-12}\\ &=\displaystyle \frac{18-12\sqrt{3}}{-3}\\ &=-6+4\sqrt{3}\\ &=4\sqrt{3}-6 \end{aligned}&h.&\begin{aligned}\displaystyle &\frac{3-\sqrt{2}}{3+\sqrt{2}}\\ &=\displaystyle \frac{3-\sqrt{2}}{3+\sqrt{2}}\times \frac{3-\sqrt{2}}{3-\sqrt{2}}\\ &=\displaystyle \frac{3^{2}-2.3.\sqrt{2}+\left ( \sqrt{2} \right )^{2}}{3^{2}-\left ( \sqrt{2} \right )^{2}}\\ &=\displaystyle \frac{9+2-6\sqrt{2}}{9-2}\\ &=\displaystyle \frac{1}{7}\left ( 11-6\sqrt{2} \right ) \end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 6.&\textrm{Sederhanakanlah pecahan berikut dengan merasionalkan penyebutnya} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle \frac{1}{\sqrt{5-2\sqrt{6}}}&b.&\displaystyle \frac{1}{\sqrt{15+\sqrt{224}}}&c.&\displaystyle \frac{1}{\sqrt{14-6\sqrt{5}}}-\frac{1}{\sqrt{14+6\sqrt{5}}}&d.&\displaystyle \frac{2}{\sqrt{7+2\sqrt{12}}}-\frac{1}{\sqrt{7+2\sqrt{12}}}\\ &&&&&&&\\ &e.&\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}&f.&\displaystyle \frac{6}{\sqrt{11}-\sqrt{6}+\sqrt{5}}&g.&\displaystyle \frac{12}{3+\sqrt{7}-\sqrt{2}}&h.&\displaystyle \frac{5}{4-2\sqrt{3}+\sqrt{5}}\\ \end{array}.

.\quad\: \, \begin{aligned}&\begin{array}{|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a.&\begin{aligned}\displaystyle \frac{1}{\sqrt{5-2\sqrt{6}}}&=\displaystyle \frac{1}{5-2\sqrt{6}}\\ &=\displaystyle \frac{1}{\sqrt{3+2-2\sqrt{3.2}}}\\ &=\displaystyle \frac{1}{\sqrt{\left ( \sqrt{3}-\sqrt{2} \right )^{2}}}\\ &=\displaystyle \frac{1}{\sqrt{3}-\sqrt{2}}\\ &=\displaystyle \frac{1}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{3}+3\sqrt{2}}{3-2}\\ &=\sqrt{3}+\sqrt{2} \end{aligned}&b.&\begin{aligned}\displaystyle &\frac{1}{\sqrt{15+\sqrt{4.8.7}}}\\ &=\displaystyle \frac{1}{\sqrt{8+7+2\sqrt{8.7}}}\\ &=\displaystyle \frac{1}{\sqrt{\left ( \sqrt{8}+\sqrt{7} \right )^{2}}}\\ &=\displaystyle \frac{1}{\sqrt{8}+\sqrt{7}}\\ &=\displaystyle \frac{1}{\sqrt{8}+\sqrt{7}}\times \frac{\sqrt{8}-\sqrt{7}}{\sqrt{8}-\sqrt{7}}\\ &=\displaystyle \frac{\sqrt{8}-\sqrt{7}}{8-7}\\ &=\sqrt{8}-\sqrt{7} \end{aligned}&c.&\begin{aligned}\displaystyle &\frac{1}{\sqrt{14-6\sqrt{5}}}-\frac{1}{\sqrt{14+6\sqrt{5}}}\\ &=\displaystyle \frac{1}{\sqrt{14-2.3\sqrt{5}}}-\frac{1}{\sqrt{14+2.3\sqrt{5}}}\\ &=\displaystyle \frac{1}{\sqrt{9+5-2\sqrt{9.5}}}-\frac{1}{\sqrt{9+5+2\sqrt{9.5}}}\\ &=\displaystyle \frac{1}{\sqrt{9}-\sqrt{5}}-\displaystyle \frac{1}{\sqrt{9}+\sqrt{5}}\\ &=\displaystyle \left ( \frac{\sqrt{9}+\sqrt{5}}{9-5} \right )-\displaystyle \left ( \frac{\sqrt{9}-\sqrt{5}}{9-5} \right )\\ &=\displaystyle \frac{1}{4}\left ( 2\sqrt{5} \right )\\ &=\displaystyle \frac{1}{2}\sqrt{5} \end{aligned} \\\hline \end{array} \end{aligned}.

Sumber Referensi

  1. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2005. Matematika untuk SMA dan MA Kelas XII Program Ilmu Alam. Jakarta: Gelora Aksara Pratama.
  2. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2007. Matematika  SMA dan MA untuk Kelas XII Semester 2 Program IPA Standar Isi 2006. Jakarta: esis.
  3. Soetiyono, Sajaka, K. A., Suprijanto, S., Marwanta, Murniati, S., dan Herynugroho. 2007. Matematika Interaktif 3B Sekolah Menengah Atas Kelas XII Program Ilmu Pengetahuan Alam. Jakarta: Yudistira.
  4. Tim IGMP Matematika SMA. ….. . Tabloid Matematika Kurikulum 2006. Semarang: CV. Sarana Ilmu.
  5. Tung, K. Y. 2012. Pintar Matematika SMA Kelas XII IPA Untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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