Lanjutan Fungsi Eksponen

$\text{C.2 Operasi Bilangan Bentuk Akar}$.

C. 2. 1  Sifat-sifat yang berlaku pada operasi bilangan bentuk akar

$\begin{array}{rl}& \\ 1.& a\sqrt[n]{c}+b\sqrt[n]{c}=(a+b)\sqrt[n]{c}\\ 2.& a\sqrt[n]{c}-b\sqrt[n]{c}=(a-b)\sqrt[n]{c}\\ 3.& \sqrt[n]{a}.\sqrt[n]{b}=\sqrt[n]{ab}\\ 4.& \sqrt[n]{a^n}=a\\ 5.& a\sqrt[n]{c}xb\sqrt[n]{d}=ab\sqrt[n]{cd}\\ 6.& \frac{a\sqrt[n]{c}}{b\sqrt[n]{d}}=\frac{a}{b}.\sqrt[n]{\frac{c}{d}}\\ 7.& \sqrt{(a+b)+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\\ 8.& \sqrt{(a+b)-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}\end{array}$.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Sederhanakanlah bentuk akar berikut}\\ & \begin{array}{llllll}\text{a}.& \sqrt{8}& \text{f}.& \sqrt[3]{16}& \text{k}.& \sqrt{8x^5},x\ge 0\\ \text{b}.& \sqrt{12}& \text{g}.& \sqrt[3]{32}& \text{l}.& \sqrt{48x^6{y}^{11}},y\ge 0\\ \text{c}.& \sqrt{27}& \text{h}.& \sqrt[3]{54}& \text{m}.& 2\sqrt{8}\times \sqrt{3}\\ \text{d}.& \sqrt{28}& \text{i}.& \sqrt[3]{81}& \text{n}.& 3\sqrt{6}\times 2\sqrt{2}\\ \text{e}.& \sqrt{32}& \text{j}.& \sqrt[3]{625}& \text{o}.& 2\sqrt[3]{6}\times 6\sqrt[3]{9}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \text{a}.& \sqrt{8}=\sqrt{4\times 2}=\sqrt{2^2}\times \sqrt{2}=2\sqrt{2}\\ \text{b}.& \sqrt{12}=\sqrt{4\times 3}=\sqrt{2^2}\times \sqrt{3}=2\sqrt{3}\\ \text{c}.& \sqrt{27}=\sqrt{9\times 3}=\sqrt{3^2}\times \sqrt{3}=3\sqrt{3}\\ \text{d}.& \sqrt{28}=\sqrt{4\times 7}=\sqrt{2^2}\times \sqrt{7}=2\sqrt{7}\\ \text{e}.& \sqrt{32}=\sqrt{16\times 2}=\sqrt{4^2}\times \sqrt{2}=4\sqrt{2}\\ \text{f}.& \sqrt[3]{16}=\sqrt[3]{8\times 2}=\sqrt[3]{2^3}\times \sqrt[3]{2}=2\sqrt[3]{2}\\ \text{g}.& \sqrt[3]{32}=\sqrt[3]{8\times 4}=\sqrt[3]{2^3}\times \sqrt[3]{4}=2\sqrt[3]{4}\\ \text{h}.& \sqrt[3]{54}=\sqrt[3]{27\times 2}=\sqrt[3]{3^3}\times \sqrt[3]{2}=3\sqrt[3]{2}\\ \text{i}.& \sqrt[3]{81}=\sqrt[3]{27\times 3}=\sqrt[3]{3^3}\times \sqrt[3]{3}=3\sqrt[3]{3}\\ \text{j}.& \sqrt[3]{625}=\sqrt[3]{125\times 5}=\sqrt[3]{5^3}\times \sqrt[3]{5}=5\sqrt[3]{5}\\ \text{k}.& \sqrt{8x^5}=\sqrt{4.2.x^4.x^1}=\sqrt{2^2}\times \sqrt{2}\times \sqrt{x^4}\times \sqrt{x}\\ & =2.\sqrt{2}.x^2.\sqrt{x}=2x^2\sqrt{2x},x\ge 0\\ \text{l}.& \sqrt{48x^6{y}^{11}}=\sqrt{16.3.x^6.y^{10}.y^1}\\ & =\sqrt{4^2}\times \sqrt{3}\times \sqrt{x^6}\times \sqrt{y^{10}}\times \sqrt{y}\\ & =4\sqrt{3}.x^3.y^5.\sqrt{y}=4x^3{y}^5\sqrt{3y},\ge 0\\ \text{m}.& 2\sqrt{8}\times \sqrt{3}=2\sqrt{4\times 2}\times \sqrt{3}\\ & =2\sqrt{2^2}\times \sqrt{2}\times \sqrt{3}=2.2.\sqrt{2.3}\\ & =4\sqrt{6}\\ \text{n}.& 3\sqrt{6}\times 2\sqrt{2}=3\sqrt{2\times 3}\times 2\sqrt{2}\\ & =3\times 2\times \sqrt{2^2\times 3}=6\times \sqrt{2^2}\times \sqrt{3}\\ & =6\times 2\times \sqrt{3}=12\sqrt{3}\sqrt{6}\\ \text{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}\\ & =12\times \sqrt[3]{{2.3}^3}=12\times \sqrt[3]{2}\times \sqrt[3]{3^3}\\ & =12\times \sqrt[3]{2}\times 3\\ & =36\sqrt[3]{2}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah pangkat rasional dari}\\ & \text{a}.\sqrt{y\sqrt[3]{x^{2}y}}\\ & \text{b}.\sqrt[3]{x^3\sqrt[5]{x^3\sqrt{x^3}}}\\ & \text{c}.\sqrt[3]{x^2\sqrt{x\sqrt[5]{x^2}}}\\ & \text{d}.xyz\sqrt[3]{{\displaystyle \frac{xy}{z^5}}}\sqrt[3]{{\displaystyle \frac{xz}{y^5}}}\sqrt[3]{{\displaystyle \frac{yz}{x^5}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \text{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}}\\ & =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}}}\\ & =x^{{.}^{\frac{1}{3}}}.y^{{.}^{\frac{2}{3}}}\\ \text{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}}}}\\ & =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}}}\\ & =x^{{.}^{1+\frac{1}{5}+\frac{1}{10}}}=x^{{.}^{\frac{10+2+1}{10}}}=x^{{.}^{\frac{13}{10}}}\\ \text{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}}}}\\ & =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}}}\\ & =x^{{.}^{\frac{20+5+2}{30}}}=x^{{.}^{\frac{27}{30}}}=x^{{.}^{\frac{9}{10}}}\\ \text{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}}}\\ & =xyz\sqrt[3]{{\displaystyle \frac{1}{x^{(5-2)}{y}^{(5-2)}{z}^{(5-2)}}}}\\ & =xyz\sqrt[3]{{\displaystyle \frac{1}{x^3{y}^3{z}^3}}}=xyz\sqrt[3]{{\displaystyle \frac{1}{(xyz{)}^3}}}\\ & =xyz.{\displaystyle \frac{1}{xyz}={\displaystyle \frac{xyz}{xyz}=1}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika}a,b\text{bilangan positif dan}\\ & \sqrt{a^{2}b\sqrt[3]{a{b}^2\sqrt{ab}}}=a^x.b^y,\text{tentukan nilai}x-y\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \sqrt{a^{2}b\sqrt[3]{a{b}^2\sqrt{ab}}}=a^x.b^y\\ & \text{perhatikan cara menguraikannya}\\ & \sqrt{a^{2}b\sqrt[3]{a{b}^2\sqrt{ab}}}=\sqrt{a^{2}b\sqrt[3]{a{b}^2.a^{{.}^{\frac{1}{2}}}{b}^{{.}^{\frac{1}{2}}}}}\\ & =\sqrt{a^{2}b\sqrt[3]{a^{{.}^{1+\frac{1}{2}}}{b}^{{.}^{2+\frac{1}{2}}}}}=\sqrt{a^{2}b\sqrt[3]{a^{{.}^{\frac{3}{2}}}{b}^{{.}^{\frac{5}{2}}}}}\\ & =\sqrt{a^{2}b.a^{{.}^{\frac{3}{2.3}}}{b}^{{.}^{\frac{5}{2.3}}}}=\sqrt{a^{2}b.a^{{.}^{\frac{1}{2}}}{b}^{{.}^{\frac{5}{6}}}}\\ & =\sqrt{a^{{.}^{2+\frac{1}{2}}}.b^{{.}^{1+\frac{5}{6}}}}=\sqrt{a^{{.}^{\frac{5}{2}}}{b}^{{.}^{\frac{11}{6}}}}=a^{{.}^{\frac{5}{2.2}}}{b}^{{.}^{\frac{11}{6.2}}}\\ & =a^{{.}^{\frac{5}{4}}}{b}^{{.}^{\frac{11}{12}}}\\ & =a^x{b}^y\\ & \text{maka}x={\displaystyle \frac{5}{4},\text{dan}y=\frac{11}{12}}\\ & x-y={\displaystyle \frac{5}{4}-\frac{11}{12}={\displaystyle \frac{15-11}{12}=\frac{4}{12}={\displaystyle \frac{1}{3}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& (\mathbf{\text{Matematika Dasar UM UGM 2008}})\\ & \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{\sqrt[6]{x^2}\sqrt[3]{x^2\sqrt{x+1}}}{x\sqrt[6]{x+1}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sqrt[6]{x^2}\sqrt[3]{x^2\sqrt{x+1}}}{x\sqrt[6]{x+1}}}\\ & ={\displaystyle \frac{\sqrt[6]{x^2}.\sqrt[3.2]{{(x^2.\sqrt{x+1})}^2}}{\sqrt[6]{x^6}.\sqrt[6]{x+1}}={\displaystyle \frac{\sqrt[6]{x^2}.\sqrt[6]{x^4.(x+1)}}{\sqrt[6]{x^6(x+1)}}}}\\ & ={\displaystyle \frac{\sqrt[6]{x^{(2+4)}(x+1)}}{\sqrt[6]{x^6(x+1)}}={\displaystyle \frac{\sqrt[6]{x^6(x+1)}}{\sqrt[6]{x^6(x+1)}}}}\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai dari}\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}=....\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\\ & \text{berikut uraiannya}\\ & \text{Misalkan}\\ & x^2=x^2,\text{maka}{x}^2=1+(x^2-1)\\ & x^2=1+(x-1)(x+1)\\ & x^2=1+(x-1)\sqrt{(x+1{)}^2}\\ & x^2=1+(x-1)\sqrt{1+((x+1{)}^2-1)}\\ & x^2=1+(x-1)\sqrt{1+(x+1-1)(x+1+1)}\\ & x^2=1+(x-1)\sqrt{1+x(x+2)}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{(x+2{)}^2}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+((x+2{)}^2-1)}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+2-1)(x+2+1)}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{(x+3{)}^2}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+((x+3{)}^2-1)}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+3-1)(x+3+1)}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{(x+4{)}^2}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+((x+4{)}^2-1)}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+4-1)(x+4+1)}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)(x+5)}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{...}}}}}\\ & x=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\cdots }}}}}\\ & \text{maka}\\ & \cdots =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\\ & \text{Jelas tampak bahwa nilai}x\text{yang memenuhi adalah}\\ & x=3\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakanlah bentuk}\\ & \text{a}.\sqrt{3+2\sqrt{2}}\text{d}.\sqrt{21-4\sqrt{5}}\\ & \text{b}.\sqrt{6-\sqrt{32}}\text{e}.\sqrt{6-2\sqrt{8}}\\ & \text{c}.\sqrt{7+4\sqrt{3}}\text{f}.\sqrt{5+\sqrt{24}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Inga}& \text{t bahwa}:\sqrt{a+b\pm 2\sqrt{ab}}=\sqrt{a}\pm \sqrt{b},a\ge b\\ \text{a}.& \sqrt{3+2\sqrt{2}}=\sqrt{2+1+2\sqrt{2.1}}=\sqrt{2}+1\\ \text{b}.& \sqrt{6-\sqrt{32}}=\sqrt{6-\sqrt{4.4.2}}=\sqrt{4+2-2\sqrt{4.2}}\\ & =\sqrt{4}-\sqrt{2}=2-\sqrt{2}\\ \text{c}.& \sqrt{7+4\sqrt{3}}=\sqrt{4+3+2.2\sqrt{3}}=\sqrt{4+3+2\sqrt{4.3}}\\ & =\sqrt{4}+\sqrt{3}=2+\sqrt{3}\\ \text{d}.& \sqrt{21-4\sqrt{5}}=\sqrt{20+1-2.2\sqrt{5}}=\sqrt{20+1-2\sqrt{4.5}}\\ & =\sqrt{20+1-2\sqrt{20.1}}=\sqrt{20}-1=\sqrt{4.5}-1=2\sqrt{5}-1\\ \text{e}.& \sqrt{6-2\sqrt{8}}=\sqrt{4+2-2\sqrt{4.2}}=\sqrt{4}-\sqrt{2}=2-\sqrt{2}\\ \text{f}.& \sqrt{5+\sqrt{24}}=\sqrt{3+2+\sqrt{4.3.2}}=\sqrt{3+2+2\sqrt{3.2}}\\ & =\sqrt{3}+\sqrt{2}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Sederhanakan bentuk berikut}\\ & \text{a}.\sqrt{0,3+\sqrt{0,08}}\\ & \text{b}.\sqrt{94+2\sqrt{2013}}\\ & \text{c}.\sqrt{17+4\sqrt{15}}=a\sqrt{3}+b\sqrt{5},\text{tentukan}b-a\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Inga}& \text{t bahwa}:\sqrt{p+q\pm 2\sqrt{pq}}=\sqrt{p}\pm \sqrt{q},p\ge q\\ \text{a}.& \sqrt{0,3+\sqrt{0,08}}=\sqrt{0,3+\sqrt{4.(0,02)}}=\sqrt{0,3+2\sqrt{0,02}}\\ & =\sqrt{0,2+0,1+2\sqrt{(0,2).(0,1)}}=\sqrt{0,2}+\sqrt{0,1}\\ \text{b}.& \sqrt{94+2\sqrt{2013}}=\sqrt{61+33+2\sqrt{61.33}}=\sqrt{61}+\sqrt{33}\\ \text{c}.& \sqrt{17+4\sqrt{15}}=\sqrt{17+2.2\sqrt{15}}=\sqrt{17+2\sqrt{4.15}}\\ & =\sqrt{17+2\sqrt{60}}=\sqrt{12+5+2\sqrt{12.5}}=\sqrt{12}+\sqrt{5}\\ & =\sqrt{4.3}+\sqrt{1.5}=2\sqrt{3}+1\sqrt{5}=a\sqrt{3}+b\sqrt{5}\{\begin{array}{ll}a& =2\\ b& =1\end{array}\\ & \text{maka}b-a=1-2=-1\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Bentuk paling sederhana dari}\\ & \sqrt[4]{49-20\sqrt{6}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \sqrt[4]{49-20\sqrt{6}}\\ & =\sqrt{\sqrt{49-2.10\sqrt{6}}}=\sqrt{\sqrt{49-2\sqrt{100.6}}}\\ & =\sqrt{\sqrt{49-2\sqrt{600}}}=\sqrt{\sqrt{25+24-2\sqrt{25.24}}}\\ & =\sqrt{\sqrt{25}-\sqrt{24}}=\sqrt{5-\sqrt{24}}=\sqrt{5-\sqrt{4.6}}\\ & =\sqrt{5-2\sqrt{6}}=\sqrt{3+2-2\sqrt{3.2}}=\sqrt{3}-\sqrt{2}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Bentuk paling sederhana dari}\\ & {\left(\sqrt{52+6\sqrt{43}}\right)}^3-{\left(\sqrt{52-6\sqrt{43}}\right)}^3\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Inga}& \text{tlah bentuk}:(A-B{)}^3=A^3-B^3-3AB(A-B)\\ & \iff A^3-B^3=(A-B{)}^3+3AB(A-B)\\ \bullet & \sqrt{52+6\sqrt{43}}=\sqrt{52+2.3\sqrt{43}}=\sqrt{52+2\sqrt{43.9}}\\ & =\sqrt{43+9+2\sqrt{43.9}}=\sqrt{43}+\sqrt{9}=\sqrt{43}+3\\ \bullet & \sqrt{52-6\sqrt{43}}=\sqrt{52-2.3\sqrt{43}}=\sqrt{52-2\sqrt{43.9}}\\ & =\sqrt{43+9-2\sqrt{43.9}}=\sqrt{43}-\sqrt{9}=\sqrt{43}-3\\ & \text{misalkan}\{\begin{array}{ll}A& =\sqrt{43}+3\\ B& =\sqrt{43}-3\end{array}\\ & A^3-B^3=(A-B{)}^3+3AB(A-B)\\ & ={(\sqrt{43}+3-(\sqrt{43}-3))}^3+3(\sqrt{43}+3)(\sqrt{43}-3)(\sqrt{43}+3-(\sqrt{43}-3))\\ & ={\left(6\right)}^3+3({\sqrt{43}}^2-3^2)\left(6\right)\\ & =216+18(43-9)\\ & =216+18.34\\ & =216+612\\ & =828\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Sederhanakanlah bentuk akar berikut}\\ & \begin{array}{ll}\text{a}.& 3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\ \text{b}.& 5\sqrt{3}+5\sqrt{27}-2\sqrt{75}\\ \text{c}.& 3\sqrt{50}-4\sqrt{32}-\sqrt{2}\\ \text{d}.& (2+\sqrt{2})(4-\sqrt{2})\\ \text{e}.& (3\sqrt{2}+\sqrt{3})(\sqrt{2}-2\sqrt{3})\\ \text{f}.& (4\sqrt{3}-3\sqrt{5})(2\sqrt{3}+\sqrt{5})\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \text{a}.& 3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\ & =3\sqrt{2}+5\sqrt{4.2}-\sqrt{16.2}\\ & =3\sqrt{2}+5.2\sqrt{2}-4\sqrt{2}\\ & =(3+10-4)\sqrt{2}=9\sqrt{2}\\ \text{b}.& 5\sqrt{3}+5\sqrt{27}-2\sqrt{75}\\ & =5\sqrt{3}+5\sqrt{9.3}-2\sqrt{25.3}\\ & =5\sqrt{3}+5.3\sqrt{3}-2.5\sqrt{3}\\ & =(5+15-10)\sqrt{3}=10\sqrt{3}\\ \text{c}.& 3\sqrt{50}-4\sqrt{32}-\sqrt{2}\\ & =3\sqrt{25.2}-4\sqrt{16.2}-\sqrt{1.2}\\ & =3.5\sqrt{2}-4.4\sqrt{2}-1\sqrt{2}\\ & =(15-16-1)\sqrt{2}=-2\sqrt{2}\\ \text{d}.& (2+\sqrt{2})(4-\sqrt{2})\\ & =2.4-2.\sqrt{2}+4.\sqrt{2}-\sqrt{2.2}\\ & =8+(4-2)\sqrt{2}-2\\ & =6+2\sqrt{2}\\ \text{e}.& (3\sqrt{2}+\sqrt{3})(\sqrt{2}-2\sqrt{3})\\ & =3\sqrt{2.2}-3.2.\sqrt{2.3}+\sqrt{3.2}-2\sqrt{3.3}\\ & =3.2-6\sqrt{6}+1\sqrt{6}-3.2\\ & =6-6+(1-6)\sqrt{6}=-5\sqrt{6}\\ \text{f}.& (4\sqrt{3}-3\sqrt{5})(2\sqrt{3}+\sqrt{5})\\ & =4.2.\sqrt{3.3}+4\sqrt{3.5}-3.2.\sqrt{5.3}-3\sqrt{5.5}\\ & =8.3+4\sqrt{15}-6\sqrt{15}-3.5\\ & =24-15+(4-6)\sqrt{15}=9-2\sqrt{15}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. 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.