$\Large\textrm{C.2 Operasi Bilangan Bentuk Akar}$.
C. 2. 1 Sifat-sifat yang berlaku pada operasi bilangan bentuk akar
$\color{blue}\begin{aligned}&\\ 1.\quad&a\sqrt[n]{c}+b\sqrt[n]{c}=\left ( a+b \right )\sqrt[n]{c}\\ 2.\quad&a\sqrt[n]{c}-b\sqrt[n]{c}=\left ( a-b \right )\sqrt[n]{c}\\ 3.\quad&\sqrt[n]{a}.\sqrt[n]{b}=\sqrt[n]{ab}\\ 4.\quad&\sqrt[n]{a^{n}}=a\\ 5.\quad&a\sqrt[n]{c} x b\sqrt[n]{d} = ab\sqrt[n]{cd}\\ 6.\quad&\frac{a\sqrt[n]{c}}{b\sqrt[n]{d}}=\frac{a}{b}.\sqrt[n]{\frac{c}{d}}\\ 7.\quad&\sqrt{\left ( a+b \right )+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\\ 8.\quad&\sqrt{\left ( a+b \right )-2\sqrt{ab}}=\sqrt{a}-\sqrt{b} \end{aligned}$.
$\LARGE\colorbox{yellow}{ CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Sederhanakanlah bentuk akar berikut}\\ &\begin{array}{lllllll} \textrm{a}.&\sqrt{8}&\textrm{f}.&\sqrt[3]{16}&\textrm{k}.&\sqrt{8x^{5}},\: \: x\geq 0\\ \textrm{b}.&\sqrt{12}&\textrm{g}.&\sqrt[3]{32}&\textrm{l}.&\sqrt{48x^{6}y^{11}},\: \: y\geq 0\\ \textrm{c}.&\sqrt{27}&\textrm{h}.&\sqrt[3]{54}&\textrm{m}.&2\sqrt{8}\times \sqrt{3}\\ \textrm{d}.&\sqrt{28}&\textrm{i}.&\sqrt[3]{81}&\textrm{n}.&3\sqrt{6}\times 2\sqrt{2}\\ \textrm{e}.&\sqrt{32}&\textrm{j}.&\sqrt[3]{625}&\textrm{o}.&2\sqrt[3]{6}\times 6\sqrt[3]{9} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{array}{lllllll}\\ \textrm{a}.&\sqrt{8}=\sqrt{4\times 2}=\sqrt{2^{2}}\times \sqrt{2}=2\sqrt{2}\\ \textrm{b}.&\sqrt{12}=\sqrt{4\times 3}=\sqrt{2^{2}}\times \sqrt{3}=2\sqrt{3}\\ \textrm{c}.&\sqrt{27}=\sqrt{9\times 3}=\sqrt{3^{2}}\times \sqrt{3}=3\sqrt{3}\\ \textrm{d}.&\sqrt{28}=\sqrt{4\times 7}=\sqrt{2^{2}}\times \sqrt{7}=2\sqrt{7}\\ \textrm{e}.&\sqrt{32}=\sqrt{16\times 2}=\sqrt{4^{2}}\times \sqrt{2}=4\sqrt{2}\\ \textrm{f}.&\sqrt[3]{16}=\sqrt[3]{8\times 2}=\sqrt[3]{2^{3}}\times \sqrt[3]{2}=2\sqrt[3]{2}\\ \textrm{g}.&\sqrt[3]{32}=\sqrt[3]{8\times 4}=\sqrt[3]{2^{3}}\times \sqrt[3]{4}=2\sqrt[3]{4}\\ \textrm{h}.&\sqrt[3]{54}=\sqrt[3]{27\times 2}=\sqrt[3]{3^{3}}\times \sqrt[3]{2}=3\sqrt[3]{2}\\ \textrm{i}.&\sqrt[3]{81}=\sqrt[3]{27\times 3}=\sqrt[3]{3^{3}}\times \sqrt[3]{3}=3\sqrt[3]{3}\\ \textrm{j}.&\sqrt[3]{625}=\sqrt[3]{125\times 5}=\sqrt[3]{5^{3}}\times \sqrt[3]{5}=5\sqrt[3]{5}\\ \textrm{k}.&\sqrt{8x^{5}}=\sqrt{4.2.x^{4}.x^{1}}=\sqrt{2^{2}}\times \sqrt{2}\times \sqrt{x^{4}}\times \sqrt{x}\\ &\quad\quad \: \: \: =2.\sqrt{2}.x^{2}.\sqrt{x}=2x^{2}\sqrt{2x},\: \: \: x\geq 0\\ \textrm{l}.&\sqrt{48x^{6}y^{11}}=\sqrt{16.3.x^{6}.y^{10}.y^{1}}\\ &\quad\quad \: \: \: =\sqrt{4^{2}}\times \sqrt{3}\times \sqrt{x^{6}}\times \sqrt{y^{10}}\times \sqrt{y}\\ &\quad\quad \: \: \: =4\sqrt{3}.x^{3}.y^{5}.\sqrt{y}=4x^{3}y^{5}\sqrt{3y},\: \: \: \geq 0\\ \textrm{m}.&2\sqrt{8}\times \sqrt{3}=2\sqrt{4\times 2}\times \sqrt{3}\\ &\quad\quad \: \: \: =2\sqrt{2^{2}}\times \sqrt{2}\times \sqrt{3}=2.2.\sqrt{2.3}\\ &\quad\quad \: \: \: =4\sqrt{6}\\ \textrm{n}.&3\sqrt{6}\times 2\sqrt{2}=3\sqrt{2\times 3}\times 2\sqrt{2}\\ &\quad\quad \: \: \: =3\times 2\times \sqrt{2^{2}\times 3}=6\times \sqrt{2^{2}}\times \sqrt{3}\\ &\quad\quad \: \: \: =6\times 2\times \sqrt{3}=12\sqrt{3}\sqrt{6}\\ \textrm{o}.&2\sqrt[3]{6}\times 6\sqrt[3]{9}=2.6.\sqrt[3]{6\times 9}=12\times \sqrt[3]{2.3.3.3}\\ &\quad\quad \: \: \: =12\times \sqrt[3]{2.3^{3}}=12\times \sqrt[3]{2}\times \sqrt[3]{3^{3}}\\ &\quad\quad \: \: \: =12\times \sqrt[3]{2}\times 3\\ &\quad\quad \: \: \: =36\sqrt[3]{2} \end{array} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah pangkat rasional dari}\\ &\textrm{a}.\quad \sqrt{y\sqrt[3]{x^{2}y}}\\ &\textrm{b}.\quad \sqrt[3]{x^{3}\sqrt[5]{x^{3}\sqrt{x^{3}}}}\\ &\textrm{c}.\quad \sqrt[3]{x^{2}\sqrt{x\sqrt[5]{x^{2}}}}\\ &\textrm{d}.\quad xyz\sqrt[3]{\displaystyle \frac{xy}{z^{5}}}\sqrt[3]{\displaystyle \frac{xz}{y^{5}}}\sqrt[3]{\displaystyle \frac{yz}{x^{5}}}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{array}{lllllll}\\ \textrm{a}.&\sqrt{y\sqrt[3]{x^{2}y}}=\sqrt{y\left ( x^{2}y \right )^{\frac{1}{3}}}=\left ( y\left ( x^{2}y \right )^{\frac{1}{3}} \right )^{\frac{1}{2}}\\ &\quad\quad \: \: \: =y^{.^{\frac{1}{2}}}.x^{.^{\frac{2}{3}.\frac{1}{2}}}.y^{.^{\frac{1}{3}.\frac{1}{2}}}=y^{.^{\frac{1}{2}+\frac{1}{6}}}x^{.^{\frac{1}{3}}}=x^{.^{\frac{1}{3}}}.y^{.^{\frac{4}{6}}}\\ &\quad\quad \: \: \: =x^{.^{\frac{1}{3}}}.y^{.^{\frac{2}{3}}}\\ \textrm{b}.&\sqrt[3]{x^{3}\sqrt[5]{x^{3}\sqrt{x^{3}}}}=\sqrt[3]{x^{3}\sqrt[5]{x^{3}.x^{.^{\frac{3}{2}}}}}=\sqrt[3]{x^{3}.x^{.^{\frac{3}{5}}}x^{.^{\frac{3}{2.5}}}}\\ &\quad\quad \: \: \: =x^{.^{\frac{3}{3}}}.x^{.^{\frac{3}{5.3}}}.x^{.^{\frac{3}{2.5.3}}}=x^{1}+x^{.^{\frac{1}{5}}}.x^{.^{\frac{1}{10}}}\\ &\quad\quad \: \: \: =x^{.^{1+\frac{1}{5}+\frac{1}{10}}}=x^{.^{\frac{10+2+1}{10}}}=x^{.^{\frac{13}{10}}}\\ \textrm{c}.&\sqrt[3]{x^{2}\sqrt{x\sqrt[5]{x^{2}}}}=\sqrt[3]{x^{2}\sqrt{x.x^{.^{\frac{2}{5}}}}}=\sqrt[3]{x^{2}.x^{.^{\frac{1}{2}}}.x^{.^{\frac{2}{5.2}}}}\\ &\quad\quad \: \: \: =x^{.^{\frac{2}{3}}}.x^{.^{\frac{1}{2.3}}}.x^{.^{\frac{2}{5.2.3}}}=x^{.^{\frac{2}{3}}}.x^{.^{\frac{1}{6}}}.x^{.^{\frac{1}{15}}}\\ &\quad\quad \: \: \: =x^{.^{\frac{20+5+2}{30}}}=x^{.^{\frac{27}{30}}}=x^{.^{\frac{9}{10}}}\\ \textrm{d}.&xyz\sqrt[3]{\displaystyle \frac{xy}{z^{5}}}\sqrt[3]{\displaystyle \frac{xz}{y^{5}}}\sqrt[3]{\displaystyle \frac{yz}{x^{5}}}=xyz\sqrt[3]{\displaystyle \frac{x^{2}y^{2}z^{2}}{x^{5}y^{5}z^{5}}}\\ &\quad\quad \: \: \: =xyz\sqrt[3]{\displaystyle \frac{1}{x^{(5-2)}y^{(5-2)}z^{(5-2)}}}\\ &\quad\quad \: \: \: =xyz\sqrt[3]{\displaystyle \frac{1}{x^{3}y^{3}z^{3}}}=xyz\sqrt[3]{\displaystyle \frac{1}{(xyz)^{3}}}\\ &\quad\quad \: \: \: =xyz.\displaystyle \frac{1}{xyz}=\displaystyle \frac{xyz}{xyz}=1 \end{array} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: a,\: b\: \: \textrm{bilangan positif dan}\\ &\sqrt{a^{2}b\sqrt[3]{ab^{2}\sqrt{ab}}}=a^{x}.b^{y},\: \: \textrm{tentukan nilai}\: \: x-y\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\sqrt{a^{2}b\sqrt[3]{ab^{2}\sqrt{ab}}}=a^{x}.b^{y}\\ &\color{red}\textrm{perhatikan cara menguraikannya}\\ &\sqrt{a^{2}b\sqrt[3]{ab^{2}\sqrt{ab}}}=\sqrt{a^{2}b\sqrt[3]{ab^{2}.a^{.^{\frac{1}{2}}}b^{.^{\frac{1}{2}}}}}\\ &=\sqrt{a^{2}b\sqrt[3]{a^{.^{1+\frac{1}{2}}}b^{.^{2+\frac{1}{2}}}}}=\sqrt{a^{2}b\sqrt[3]{a^{.^{\frac{3}{2}}}b^{.^{\frac{5}{2}}}}}\\ &=\sqrt{a^{2}b.a^{.^{\frac{3}{2.3}}}b^{.^{\frac{5}{2.3}}}}=\sqrt{a^{2}b.a^{.^{\frac{1}{2}}}b^{.^{\frac{5}{6}}}}\\ &=\sqrt{a^{.^{2+\frac{1}{2}}}.b^{.^{1+\frac{5}{6}}}}=\sqrt{a^{.^{\frac{5}{2}}}b^{.^{\frac{11}{6}}}}=a^{.^{\frac{5}{2.2}}}b^{.^{\frac{11}{6.2}}}\\ &=a^{.^{\frac{5}{4}}}b^{.^{\frac{11}{12}}}\\ &=a^{x}b^{y}\\ &\quad \textrm{maka}\: \: x=\displaystyle \frac{5}{4},\: \: \textrm{dan}\: \: y=\frac{11}{12}\\ &\quad x-y=\displaystyle \frac{5}{4}-\frac{11}{12}=\displaystyle \frac{15-11}{12}=\frac{4}{12}=\color{blue}\displaystyle \frac{1}{3} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&(\textbf{Matematika Dasar UM UGM 2008})\\ &\textrm{Bentuk sederhana dari}\\ &\qquad\qquad \displaystyle \frac{\sqrt[6]{x^{2}}\sqrt[3]{x^{2}\sqrt{x+1}}}{x\sqrt[6]{x+1}}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sqrt[6]{x^{2}}\sqrt[3]{x^{2}\sqrt{x+1}}}{x\sqrt[6]{x+1}}\\ &=\displaystyle \frac{\sqrt[6]{x^{2}}.\sqrt[3.2]{\left (x^{2}.\sqrt{x+1} \right )^{2}}}{\sqrt[6]{x^{6}}.\sqrt[6]{x+1}}=\displaystyle \frac{\sqrt[6]{x^{2}}.\sqrt[6]{x^{4}.(x+1)}}{\sqrt[6]{x^{6}(x+1)}}\\ &=\displaystyle \frac{\sqrt[6]{x^{(2+4)}(x+1)}}{\sqrt[6]{x^{6}(x+1)}}=\displaystyle \frac{\sqrt[6]{x^{6}(x+1)}}{\sqrt[6]{x^{6}(x+1)}}\\ &=\color{blue}1 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 5.&\textrm{Nilai dari}\: \: \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}=....\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\\ &\textrm{berikut uraiannya}\\ &\color{red}\textrm{Misalkan}\\ &x^{2}=x^{2},\quad \textrm{maka}\: \: \: x^{2}=1+\left ( x^{2}-1 \right )\\ &x^{2}=1+(x-1)(x+1)\\ &x^{2}=1+(x-1)\sqrt{(x+1)^{2}}\\ &x^{2}=1+(x-1)\sqrt{1+((x+1)^{2}-1)}\\ &x^{2}=1+(x-1)\sqrt{1+(x+1-1)(x+1+1)}\\ &x^{2}=1+(x-1)\sqrt{1+x(x+2)}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{(x+2)^{2}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+((x+2)^{2}-1)}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+2-1)(x+2+1)}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{(x+3)^{2}}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+((x+3)^{2}-1)}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+3-1)(x+3+1)}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{(x+4)^{2}}}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+((x+4)^{2}-1)}}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+4-1)(x+4+1)}}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)(x+5)}}}}\\ &x^{2}=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{...}}}}}\\ &x=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\cdots }}}}}\\ &\textrm{maka}\\ &\cdots \: =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\\ &\textrm{Jelas tampak bahwa nilai}\: \: x\: \: \textrm{yang memenuhi adalah}\\ &\color{blue}x=3 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 6.&\textrm{Sederhanakanlah bentuk}\\ &\textrm{a}.\quad \sqrt{3+2\sqrt{2}}\qquad\textrm{d}.\quad \sqrt{21-4\sqrt{5}}\\ &\textrm{b}.\quad \sqrt{6-\sqrt{32}}\qquad\textrm{e}.\quad \sqrt{6-2\sqrt{8}}\\ &\textrm{c}.\quad \sqrt{7+4\sqrt{3}}\qquad\textrm{f}.\quad \sqrt{5+\sqrt{24}}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Inga}&\textrm{t bahwa}:\quad \sqrt{a+b\pm 2\sqrt{ab}}=\sqrt{a}\pm \sqrt{b},\quad a\geq b\\ \textrm{a}.\quad&\sqrt{3+2\sqrt{2}}=\sqrt{2+1+2\sqrt{2.1}}=\sqrt{2}+1\\ \textrm{b}.\quad&\sqrt{6-\sqrt{32}}=\sqrt{6-\sqrt{4.4.2}}=\sqrt{4+2-2\sqrt{4.2}}\\ &=\sqrt{4}-\sqrt{2}=2-\sqrt{2}\\ \textrm{c}.\quad&\sqrt{7+4\sqrt{3}}=\sqrt{4+3+2.2\sqrt{3}}=\sqrt{4+3+2\sqrt{4.3}}\\ &=\sqrt{4}+\sqrt{3}=2+\sqrt{3}\\ \textrm{d}.\quad&\sqrt{21-4\sqrt{5}}=\sqrt{20+1-2.2\sqrt{5}}=\sqrt{20+1-2\sqrt{4.5}}\\ &=\sqrt{20+1-2\sqrt{20.1}}=\sqrt{20}-1=\sqrt{4.5}-1=2\sqrt{5}-1\\ \textrm{e}.\quad&\sqrt{6-2\sqrt{8}}=\sqrt{4+2-2\sqrt{4.2}}=\sqrt{4}-\sqrt{2}=2-\sqrt{2}\\ \textrm{f}.\quad&\sqrt{5+\sqrt{24}}=\sqrt{3+2+\sqrt{4.3.2}}=\sqrt{3+2+2\sqrt{3.2}}\\ &=\sqrt{3}+\sqrt{2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 7.&\textrm{Sederhanakan bentuk berikut}\\ &\textrm{a}.\quad \sqrt{0,3+\sqrt{0,08}}\\ &\textrm{b}.\quad \sqrt{94+2\sqrt{2013}}\\ &\textrm{c}.\quad \sqrt{17+4\sqrt{15}}=a\sqrt{3}+b\sqrt{5},\: \: \textrm{tentukan}\: \: b-a\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Inga}&\textrm{t bahwa}:\quad \sqrt{p+q\pm 2\sqrt{pq}}=\sqrt{p}\pm \sqrt{q},\quad p\geq q\\ \textrm{a}.\quad&\sqrt{0,3+\sqrt{0,08}}=\sqrt{0,3+\sqrt{4.(0,02)}}=\sqrt{0,3+2\sqrt{0,02}}\\ &=\sqrt{0,2+0,1+2\sqrt{(0,2).(0,1)}}=\sqrt{0,2}+\sqrt{0,1}\\ \textrm{b}.\quad&\sqrt{94+2\sqrt{2013}}=\sqrt{61+33+2\sqrt{61.33}}=\sqrt{61}+\sqrt{33}\\ \textrm{c}.\quad&\sqrt{17+4\sqrt{15}}=\sqrt{17+2.2\sqrt{15}}=\sqrt{17+2\sqrt{4.15}}\\ &=\sqrt{17+2\sqrt{60}}=\sqrt{12+5+2\sqrt{12.5}}=\sqrt{12}+\sqrt{5}\\ &=\sqrt{4.3}+\sqrt{1.5}=2\sqrt{3}+1\sqrt{5}=a\sqrt{3}+b\sqrt{5}\quad\color{red}\begin{cases} a & =2 \\ b & = 1 \end{cases}\\ &\textrm{maka}\quad b-a=1-2=\color{blue}-1 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 8.&\textrm{Bentuk paling sederhana dari}\\ &\qquad\qquad \sqrt[4]{49-20\sqrt{6}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\sqrt[4]{49-20\sqrt{6}}\\ &=\sqrt{\sqrt{49-2.10\sqrt{6}}}=\sqrt{\sqrt{49-2\sqrt{100.6}}}\\ &=\sqrt{\sqrt{49-2\sqrt{600}}}=\sqrt{\sqrt{25+24-2\sqrt{25.24}}}\\ &=\sqrt{\sqrt{25}-\sqrt{24}}=\sqrt{5-\sqrt{24}}=\sqrt{5-\sqrt{4.6}}\\ &=\sqrt{5-2\sqrt{6}}=\sqrt{3+2-2\sqrt{3.2}}=\color{blue}\sqrt{3}-\sqrt{2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 9.&\textrm{Bentuk paling sederhana dari}\\ &\qquad\qquad \left (\sqrt{52+6\sqrt{43}} \right )^{3}-\left (\sqrt{52-6\sqrt{43}} \right )^{3}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Inga}&\textrm{tlah bentuk}:\quad \color{red}(A-B)^{3}=A^{3}-B^{3}-3AB(A-B)\\ &\qquad\qquad\qquad \Leftrightarrow \color{red}A^{3}-B^{3}=(A-B)^{3}+3AB(A-B)\\ \bullet \: \: &\sqrt{52+6\sqrt{43}}=\sqrt{52+2.3\sqrt{43}}=\sqrt{52+2\sqrt{43.9}}\\ &=\sqrt{43+9+2\sqrt{43.9}}=\sqrt{43}+\sqrt{9}=\sqrt{43}+3\\ \bullet \: \: &\sqrt{52-6\sqrt{43}}=\sqrt{52-2.3\sqrt{43}}=\sqrt{52-2\sqrt{43.9}}\\ &=\sqrt{43+9-2\sqrt{43.9}}=\sqrt{43}-\sqrt{9}=\sqrt{43}-3\\ &\textrm{misalkan}\: \: \: \begin{cases} A & =\sqrt{43}+3 \\ B & =\sqrt{43}-3 \end{cases}\\ &\color{red}A^{3}-B^{3}=(A-B)^{3}+3AB(A-B)\\ &=\left (\sqrt{43}+3 -\left (\sqrt{43}-3 \right ) \right )^{3}+3(\sqrt{43}+3)(\sqrt{43}-3)(\sqrt{43}+3-\left (\sqrt{43}-3 \right )) \\ &=\left ( 6 \right )^{3}+3\left ( \sqrt{43}^{2}-3^{2} \right )\left ( 6 \right )\\ &=216+18\left ( 43-9 \right )\\ &=216+18.34\\ &=216+612\\ &=\color{blue}828 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 10.&\textrm{Sederhanakanlah bentuk akar berikut}\\ &\begin{array}{lllllll} \textrm{a}.&3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\ \textrm{b}.&5\sqrt{3}+5\sqrt{27}-2\sqrt{75}\\ \textrm{c}.&3\sqrt{50}-4\sqrt{32}-\sqrt{2}\\ \textrm{d}.&\left ( 2+\sqrt{2} \right )\left ( 4-\sqrt{2} \right )\\ \textrm{e}.&\left ( 3\sqrt{2}+\sqrt{3} \right )\left ( \sqrt{2}-2\sqrt{3} \right )\\ \textrm{f}.&\left ( 4\sqrt{3}-3\sqrt{5} \right )\left ( 2\sqrt{3}+\sqrt{5} \right ) \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{array}{lllllll}\\ \textrm{a}.&3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\ &\quad\quad \: \: \: =3\sqrt{2}+5\sqrt{4.2}-\sqrt{16.2}\\ &\quad\quad \: \: \: =3\sqrt{2}+5.2\sqrt{2}-4\sqrt{2}\\ &\quad\quad \: \: \: =(3+10-4)\sqrt{2}=9\sqrt{2}\\ \textrm{b}.&5\sqrt{3}+5\sqrt{27}-2\sqrt{75}\\ &\quad\quad \: \: \: =5\sqrt{3}+5\sqrt{9.3}-2\sqrt{25.3}\\ &\quad\quad \: \: \: =5\sqrt{3}+5.3\sqrt{3}-2.5\sqrt{3}\\ &\quad\quad \: \: \: =(5+15-10)\sqrt{3}=10\sqrt{3}\\ \textrm{c}.&3\sqrt{50}-4\sqrt{32}-\sqrt{2}\\ &\quad\quad \: \: \: =3\sqrt{25.2}-4\sqrt{16.2}-\sqrt{1.2}\\ &\quad\quad \: \: \: =3.5\sqrt{2}-4.4\sqrt{2}-1\sqrt{2}\\ &\quad\quad \: \: \: =(15-16-1)\sqrt{2}=-2\sqrt{2}\\ \textrm{d}.&\left ( 2+\sqrt{2} \right )\left ( 4-\sqrt{2} \right )\\ &\quad\quad =2.4-2.\sqrt{2}+4.\sqrt{2}-\sqrt{2.2}\\ &\quad\quad =8+(4-2)\sqrt{2}-2\\ &\quad\quad =6+2\sqrt{2}\\ \textrm{e}.&\left ( 3\sqrt{2}+\sqrt{3} \right )\left ( \sqrt{2}-2\sqrt{3} \right )\\ &\quad\quad =3\sqrt{2.2}-3.2.\sqrt{2.3}+\sqrt{3.2}-2\sqrt{3.3}\\ &\quad\quad =3.2-6\sqrt{6}+1\sqrt{6}-3.2\\ &\quad\quad = 6-6+(1-6)\sqrt{6}=-5\sqrt{6}\\ \textrm{f}.&\left ( 4\sqrt{3}-3\sqrt{5} \right )\left ( 2\sqrt{3}+\sqrt{5} \right )\\ &\quad\quad =4.2.\sqrt{3.3}+4\sqrt{3.5}-3.2.\sqrt{5.3}-3\sqrt{5.5}\\ &\quad\quad =8.3+4\sqrt{15}-6\sqrt{15}-3.5\\ &\quad\quad =24-15+(4-6)\sqrt{15}=9-2\sqrt{15} \end{array} \end{array}$
DAFTAR PUSTAKA
- Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
- Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
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