EKSPONEN (LANJUTAN 3)

D. PERSAMAAN EKSPONEN

Berikut bentuk persamaan eksponen yang sering digunakan terangkum dalam tabel berikut beserta cara penyelesaiannya

$\begin{array}{|c|l|l|}\hline \textbf{No}&\textbf{Persamaan Eksponen}&\textbf{Penyelesaian}\\\hline 1&a^{f(x)}=1,\: \: a>0,a\neq 1&f(x)=0\\\hline 2&a^{f(x)}=a^{p},\: \: a>0,a\neq 1&f(x)=p\\\hline 3&a^{f(x)}=a^{g(x)},\: \: a>0,a\neq 1&f(x)=g(x)\\\hline 4&a^{f(x)}=b^{f(x)},\: \: a>0,a\neq 1&f(x)=0\\ &\qquad\quad \textrm{dan}\: \: b>0,\: b\neq 1&\\\hline 5&h(x)^{f(x)}=h(x)^{g(x)}&\begin{aligned}(1)\: &f(x)=g(x)\\ (2)\: &h(x)=1\\ (3)\: &h(x)=0\\ &\textrm{dengan syarat}\\ &f(x)> 0\: \: \textrm{dan}\\ &g(x)> 0\\ (4)\: &h(x)=-1\\ &\textrm{dengan syarat}\\ &f(x)\: \textrm{dan}\: g(x)\\ &\textrm{keduanya}\\ &\textrm{genap atau}\\ &\textrm{keduanya}\\ &\textrm{ganjil}\\ &\textrm{atau}\\ &\textrm{dapat juga}\\ &\textrm{ditunjukkan}\\ &(-1)^{f(x)}=(-1)^{g(x)} \end{aligned}\\\hline 6&g(x)^{f(x)}=h(x)^{f(x)}&\begin{aligned}(1)\: &g(x)=h(x)\\ (2)\: &f(x)=0\\ &\textrm{dengan syarat}\\ &g(x)\neq 0\: \: \textrm{dan}\\ &h(x)\neq 0\\ \end{aligned}\\\hline 7&f(x)^{g(x)}=1&\begin{aligned}(1)\: &f(x)=1\\ (2)\: &f(x)=-1\\ &\textrm{dengan syarat}\\ &g(x)\: \: \textrm{genap}\\ (3)\: &g(x)=0\\ &\textrm{dengan syarat}\\ &f(x)\neq 0 \end{aligned}\\\hline 8&A\left ( a^{f(x)} \right )^{2}+B\left ( a^{f(x)} \right )+C=0&\begin{aligned}&\textrm{ubah}\: \: a^{f(x)}=y\\ &\textrm{sehingga}\\ &Ay^{2}+By+C=0\\ &\textrm{selanjutnya}\\ &\textrm{substitusikan}\\ &\textrm{nilai}\: \: y\: \: \textrm{ke}\\ &\textrm{persamaan}\\ &a^{f(x)}=y \end{aligned}\\\hline \end{array}$.

$\LARGE{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukan himpunan penyelesaian dari}\\ &\textrm{a}.\quad 2^{2x-2021}=1\\ &\textrm{b}.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}=1\\ &\textrm{c}.\quad \sqrt{2}^{2x-2021}=1\\\\  &\textbf{Jawab}:\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}\\\hline \begin{aligned} 2^{2x-2021}&=1\\ 2^{2x-2021}&=2^{0}\\ 2x-2021&=0\\ 2x&=2021\\ x&=\displaystyle \frac{2021}{2}\\ & \end{aligned}&\begin{aligned} \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=1\\ \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=\left ( \frac{1}{2} \right )^{0}\\ 2x-2021&=0\\ 2x&=2021\\ x&=\displaystyle \frac{2021}{2} \end{aligned}&\begin{aligned} \sqrt{2}^{2x-2021}&=1\\ \sqrt{2}^{2x-2021}&=\sqrt{2}^{0}\\ 2x-2021&=0\\ 2x&=2021\\ x&=\displaystyle \frac{2021}{2}\\ & \end{aligned}\\\hline \textbf{HP}=\left \{ \displaystyle \frac{2021}{2} \right \}&\textbf{HP}=\left \{ \displaystyle \frac{2021}{2} \right \}&\textbf{HP}=\left \{ \displaystyle \frac{2021}{2} \right \}\\\hline \end{array}\\  \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukan himpunan penyelesaian dari}\\ &\textrm{a}.\quad 2^{2x-2021}=128\\ &\textrm{b}.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}=128\\ &\textrm{c}.\quad \sqrt{2}^{2x-2021}=128\\\\  &\textbf{Jawab}:\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}\\\hline \begin{aligned} 2^{2x-2021}&=128\\ 2^{2x-2021}&=2^{7}\\ 2x-2021&=7\\ 2x=7&+2021\\ x=\displaystyle \frac{2028}{2}&=1014\\ & \end{aligned}&\begin{aligned} \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=128\\ \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=\left ( \displaystyle \frac{1}{2} \right )^{-7}\\ 2x-2021&=-7\\ 2x=2021&-7\\ x=\displaystyle \frac{2014}{2}&=1007 \end{aligned}&\begin{aligned} \sqrt{2}^{2x-2021}&=128\\ \sqrt{2}^{2x-2021}&=\sqrt{2}^{256}\\ 2x-2021&=256\\ 2x=2021&+256\\ x&=\displaystyle \frac{2277}{2}\\ & \end{aligned}\\\hline \textbf{HP}=\left \{ 1014 \right \}&\textbf{HP}=\left \{ \displaystyle 1007 \right \}&\textbf{HP}=\left \{ \displaystyle \frac{2277}{2} \right \}\\\hline \end{array}\\  \end{array}$.

$\begin{array}{ll}\\ 3.&(\textbf{SPMB 04})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{27}{3^{2x-1}}=81^{-0,125} \: \: \textrm{adalah... .}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\displaystyle \frac{27}{3^{2x-1}}&=81^{-0,125}\\ 3^{3-(2x-1)}&=3^{4(\frac{1}{8})}\\ 3-2x+1&=-\displaystyle \frac{1}{2}\\ -2x+4&=-\displaystyle \frac{1}{2}\\ -x+2&=-\displaystyle \frac{1}{4}\\ -x&=-2-\displaystyle \frac{1}{4}\\ -x&=-2\displaystyle \frac{1}{4}\\ x&=2\displaystyle \frac{1}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&(\textbf{UMPTN 00})\\ &\textrm{Bentuk}\: \: \left (\sqrt[3]{\displaystyle \frac{1}{243}} \right )^{3x}=\left ( \displaystyle \frac{3}{3^{x-2}} \right )^{2}\sqrt[3]{\displaystyle \frac{1}{9}}\\ &\textrm{Jika}\: \: x_{0}\: \: \textrm{memenuhi persamaan, maka nilai}\\ &1-\displaystyle \frac{3}{4}x_{0}=\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\left (\sqrt[3]{\displaystyle \frac{1}{243}} \right )^{3x}&=\left ( \displaystyle \frac{3}{3^{x-2}} \right )^{2}\sqrt[3]{\displaystyle \frac{1}{9}}\\ 3^{-5x}&=3^{2(1-(x-2))}.3^{-\frac{2}{3}}\\ -5x&=2(1-(x-2))+\left ( -\displaystyle \frac{2}{3} \right ),\: \: \textrm{dikali}\: \: 3\\ -15x&=6(3-x)+(-2)\\ -15x&=18-6x-2\\ 6x-15x&=16\\ -9x&=16\\ x&=\displaystyle \frac{16}{-9}\\ x_{0}&=-\displaystyle \frac{16}{9},\: \: \textrm{selanjutnya}\\ 1-\displaystyle \frac{3}{4}x_{0}&=1-\displaystyle \frac{3}{4}\times \left (-\frac{16}{9} \right )\\ &=1+\frac{4}{3}\\ &=1+1\displaystyle \frac{1}{3}\\ &=2\displaystyle \frac{1}{3} \end{aligned} \end{array}$.

$\begin{array}{l}\\ 5.&\textrm{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Jumlah akar-akar persamaan}\\ &2023^{x^{2}-7x+7}=2024^{x^{2}-7x+7}\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}2023^{x^{2}-7x+7}&=2024^{x^{2}-7x+7}\\ \textrm{Karena basis}&\: \textrm{tidak sama},\\ \textrm{maka harusl}&\textrm{ah pangkatnya}=0,\\ x^{2}-7x+7&=0\\ \textrm{dan jumlah}\: &\textrm{akar-akarnya adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}, \: \: \textrm{dari persamaan}\\ x^{2}-7x+7&=0\begin{cases} a &=1 \\ b &=-7 \\ c &=7 \end{cases}\\ \textrm{maka}\: \: x_{1}+x_{2}&=-\displaystyle \frac{b}{a}=-\frac{-7}{1}=7 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tentukan himpunan penyelesaian dari}\\ &(x-2)^{x^{2}-7x+6}=1\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bentuk}\: \: \: f(x)^{g(x)}=1\begin{cases} f(x) & =x-2 \\  g(x) & =x^{2}-7x+5 \end{cases}\\ &\begin{array}{|l|l|l|}\hline f(x)=1&f(x)=-1&g(x)=0\\ &\textrm{Syarat}\: \: g(x)\: \: \textrm{genap}&\textrm{Syarat}\: \: f(x)\neq 0\\\hline \begin{aligned}x&-2=1\\ x&=3\\ & \end{aligned}&\begin{aligned}x&-2=-1\\ x&=2-1=1\\ & \end{aligned}&\begin{aligned}x^{2}&-7x+6=0\\ \Leftrightarrow &(x-1)(x-6)\\ \Leftrightarrow &\: x=1\: \: \textrm{atau}\: \: x=6 \end{aligned}\\\hline &\begin{aligned}\textrm{S}&\textrm{yaratnya}\: \: x\\ \textrm{u}&\textrm{ntuk}\: \: x=1\\ g&(1)=1^{2}-7+6\\ &=0\: \: (\textrm{memenuhi}) \end{aligned}&\begin{aligned}f(1)&=1-2=-1\neq 0\\ f(6)&=6-2=4\neq 0\\ &\\ & \end{aligned}\\ &\textbf{Catatan}:\: 0&\\ &\textrm{paritasnya genap}&\\\hline \end{array}\\ &\textbf{HP}=\left \{ 1,3,6 \right \} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Tentukan himpunan penyelesaian dari}\\ &(x^{2}-9x+19)^{2x+3}=(x^{2}-9x+19)^{x-1}\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bentuk}\: \: \: h(x)^{f(x)}=h(x)^{g(x)}\begin{cases} h(x) & =x^{2}-9x+19 \\  f(x) & =2x+3\\ g(x)&=x-1 \end{cases}\\ &\begin{array}{|l|l|}\hline \begin{aligned}&\textrm{Syarat-syaratnya}\\ &\bullet \: \: f(x)=g(x)\\ &\Leftrightarrow 2x+3=x-1\\ &\Leftrightarrow x=-4\\ &\bullet \: \: h(x)=1\\ &\Leftrightarrow x^{2}-9x+19=1\\ &\Leftrightarrow x^{2}-9x+18=0\\ &\Leftrightarrow (x-3)(x-6)=0\\ &\Leftrightarrow x=3\: \: \textrm{atau}\: \: x=6\\ &\bullet \: \: h(x)=0\\ &\Leftrightarrow x^{2}-9x+19=0\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{9\pm \sqrt{5}}{2}\\ &\quad \textrm{gunakan rumus ABC}\\ &\textrm{Setelah diuji keduanya}\\ &\textrm{positif, maka}\\ &x=\displaystyle \frac{9\pm \sqrt{5}}{2}\: \: \textrm{merupakan}\\ &\textbf{penyelesaian} \end{aligned} &\begin{aligned}&\textrm{lanjutannya}\\ &\bullet \: \: h(x)=-1\\ &\Leftrightarrow x^{2}-9x+19=-1\\ &\Leftrightarrow x^{2}-9x+20=0\\ &\Leftrightarrow (x-4)(x-5)=0\\ &\Leftrightarrow x=4\: \: \textrm{atau}\: \: x=5\\ &\textrm{Uji nilanya}\\ &\textrm{untuk}\: \: x=4\\ &\blacklozenge \: \: f(4)=2(4)+3\: \: \textrm{ganjil}\\ &\blacklozenge \: \: g(4)=4-1\: \: \textrm{ganjil}\\ &\textrm{karena}\: f(4),g(4)\: \textrm{keduanya }\\ &\textrm{ganjil, maka}\: \: x=4\\ &\textrm{adalah}\: \textbf{penyelesaian} \\ &\textrm{untuk}\: \: x=5\\ &\blacklozenge \: \: f(5)=2(5)+3\: \: \textrm{ganjil}\\ &\blacklozenge \: \: g(5)=5-1\: \: \textrm{genapl}\\ &\textrm{karena}\: f(4)\neq g(4),\: \textrm{maka}\: \: x=5\\ &\textrm{adalah}\: \textbf{bukan penyelesaian}\\ &\\  \end{aligned}\\\hline \end{array}\\ &\textbf{HP}=\left \{ -4,3,4,6,\displaystyle \frac{9-\sqrt{5}}{2},\frac{9+\sqrt{5}}{2} \right \} \end{array}$.

DAFTAR PUSTAKA

1. Kurnia, N, dkk. 2016. Jelajah Matematika I SMA Kelas X Peminatan MIPA. Jakarta: YUDHISTIRA.

EKSPONEN (LANJUTAN 2)

 C. 2. 2  Merasionalkan penyebut

Jika suatu pecahan penyebutnya mengandung bilangan irasional atau bentuk akar, maka penyebut ini dapat dibuat menjadi bilangan rasional. Perhatikanlah langkah berikut

$\begin{aligned}1.\quad&\displaystyle \frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{\left ( \sqrt{b^{2}} \right )}=\frac{a}{b}\sqrt{b}\\ 2.\quad&\displaystyle \frac{a}{\sqrt[3]{b}}=\frac{a}{\sqrt[3]{b}}\times \frac{\sqrt[3]{b^{2}}}{\sqrt[3]{b^{2}}}=\frac{a\sqrt[3]{b^{2}}}{\left ( \sqrt[3]{b^{3}} \right )}=\frac{a}{b}\sqrt[3]{b^{2}}\\ 3.\quad&\displaystyle \frac{a}{\sqrt[5]{b^{3}}}=\displaystyle \frac{a}{\sqrt[5]{b^{3}}}\times \frac{\sqrt[5]{b^{2}}}{\sqrt[5]{b^{2}}}=\frac{a\sqrt[5]{b^{2}}}{\sqrt[5]{b^{5}}}=\frac{a}{b}\sqrt[5]{b^{2}} \end{aligned}$

Merasionalkan di atas adalah contoh bebrapa contoh model merasionalkan jika berjenis tunggal tetapi jika nanti jenisnya lebih dari itu, maka perhatikanlah simulasi contoh berikut

$\begin{aligned}&\\ 1.\quad&\displaystyle \frac{c}{a+\sqrt{b}}=\frac{c}{a+\sqrt{b}}.\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{c\left ( a-\sqrt{b} \right )}{a^{2}-b}\\ 2.\quad&\displaystyle \frac{c}{a-\sqrt{b}}=\frac{c}{a-\sqrt{b}}.\frac{a+\sqrt{b}}{a+\sqrt{b}}=\frac{c\left ( a+\sqrt{b} \right )}{a^{2}-b}\\ 3.\quad&\displaystyle \frac{c}{\sqrt{a}+\sqrt{b}}=\frac{c}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{c\left ( \sqrt{a}-\sqrt{b} \right )}{a-b}\\ \end{aligned}$

Perhatikanlah simulasi contoh di atas, bentuk $a+\sqrt{b}$ memiliki bentuk sekawan (irasional juga) $a-\sqrt{b}$, demikian juga bentuk $\sqrt{a}+\sqrt{b}$ memiliki sekawan $\sqrt{a}-\sqrt{b}$. Disamping itu ada bentuk khusus yatu bentuk  $\sqrt[3]{a}+\sqrt[3]{b}$ memiliki bentuk sekawan $\sqrt[3]{a^{2}}-\sqrt[3]{ab}+\sqrt[3]{b^{2}}$.

$\LARGE{ CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Rasionalkanlah penyebut pecahan berikut}\\ &\textrm{dan serderhankanlah hasilnya}\\ &\textrm{a}.\quad \displaystyle \frac{2}{\sqrt{5}}\qquad\qquad \textrm{d}.\quad \displaystyle \frac{\sqrt{2}}{\sqrt{5}}\\ &\textrm{b}.\quad \displaystyle \frac{2}{5\sqrt{2}}\: \: \: \quad\quad\quad \textrm{e}.\quad \displaystyle \frac{p}{\sqrt{q}}\\ &\textrm{c}.\quad \displaystyle \frac{6}{3\sqrt{5}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \frac{2}{\sqrt{5}}=\displaystyle \frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\displaystyle \frac{2\sqrt{5}}{\sqrt{25}}=\displaystyle \frac{2}{5}\sqrt{5}\\ \textrm{b}.\quad&\displaystyle \frac{2}{5\sqrt{2}}=\displaystyle \frac{2}{5\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{2\sqrt{2}}{5\sqrt{4}}=\frac{2\sqrt{2}}{5.2}=\displaystyle \frac{1}{5}\sqrt{2} \\ \textrm{c}.\quad&\displaystyle \frac{6}{3\sqrt{5}}=\displaystyle \frac{6}{3\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\displaystyle \frac{6\sqrt{5}}{3\sqrt{25}}=\frac{6\sqrt{5}}{3.5}=\displaystyle \frac{2}{5}\sqrt{5}\\ \textrm{d}.\quad &\displaystyle \frac{\sqrt{2}}{\sqrt{5}}=\displaystyle \frac{\sqrt{2}}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{10}}{\sqrt{25}}=\frac{\sqrt{10}}{5} =\displaystyle \frac{1}{5}\sqrt{10}\\ \textrm{e}.\quad &\displaystyle \frac{p}{\sqrt{q}}=\displaystyle \frac{p}{\sqrt{q}}\times \frac{\sqrt{q}}{\sqrt{q}}=\displaystyle \frac{p\sqrt{q}}{\sqrt{q^{2}}}=\frac{p\sqrt{q}}{q}=\displaystyle \frac{p}{q}\sqrt{q} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Rasionalkanlah penyebut pecahan berikut}\\ &\textrm{dan serderhankanlah hasilnya}\\ &\textrm{a}.\quad \displaystyle \frac{3}{6-\sqrt{5}}\qquad\qquad \textrm{f}.\quad \displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}\\ &\textrm{b}.\quad \displaystyle \frac{3}{6+\sqrt{5}}\quad \quad\quad\quad \textrm{g}.\quad \frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}\\ &\textrm{c}.\quad \displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}\qquad\qquad\textrm{h}.\quad \frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}\\ &\textrm{d}.\quad \displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}\qquad\qquad\textrm{i}.\quad \frac{\sqrt{3}}{\sqrt{6-2\sqrt{5}}}\\ &\textrm{e}.\quad \displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}\: \qquad\quad\textrm{j}.\quad \frac{\sqrt{3}}{\sqrt{6+2\sqrt{5}}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \frac{3}{6-\sqrt{5}}=\displaystyle \frac{3}{6-\sqrt{5}}\times \frac{6+\sqrt{5}}{6+\sqrt{5}}=\displaystyle \frac{3\left ( 6+\sqrt{5} \right )}{6^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{18+3\sqrt{5}}{36-5}=\frac{18+3\sqrt{5}}{31}=\displaystyle \frac{1}{31}\left ( 18+3\sqrt{5} \right )\\ \textrm{b}.\quad & \displaystyle \frac{3}{6+\sqrt{5}}=\displaystyle \frac{3}{6+\sqrt{5}}\times \frac{6-\sqrt{5}}{6-\sqrt{5}}=\displaystyle \frac{3\left ( 6-\sqrt{5} \right )}{6^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{18-3\sqrt{5}}{36-5}=\frac{18-3\sqrt{5}}{31}=\displaystyle \frac{1}{31}\left ( 18-3\sqrt{5} \right )\\ \textrm{c}.\quad &\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}=\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}\times \frac{6+\sqrt{5}}{6+\sqrt{5}}=\displaystyle \frac{\sqrt{3}\left ( 6+\sqrt{5} \right )}{6^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{6\sqrt{3}+\sqrt{15}}{36-5}=\frac{6\sqrt{3}+\sqrt{15}}{31}=\displaystyle \frac{1}{31}\left ( 6\sqrt{3}+\sqrt{15} \right )\\ \textrm{d}.\quad &\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}=\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}\times \frac{6-\sqrt{5}}{6-\sqrt{5}}=\displaystyle \frac{\sqrt{3}\left ( 6-\sqrt{5} \right )}{6^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{6\sqrt{3}-\sqrt{15}}{36-5}=\frac{6\sqrt{3}-\sqrt{15}}{31}=\displaystyle \frac{1}{31}\left ( 6\sqrt{3}-\sqrt{15} \right )\\ \textrm{e}.\quad &\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}=\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\displaystyle \frac{3\left ( \sqrt{6}+\sqrt{5} \right )}{\sqrt{6}^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{3\left ( \sqrt{6}+\sqrt{5} \right )}{6-5}=\frac{3\left ( \sqrt{6}+\sqrt{5} \right )}{1}=3\left ( \sqrt{6}+\sqrt{5} \right ) \\ \textrm{f}.\quad &\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}=\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}=\displaystyle \frac{3\left ( \sqrt{6}-\sqrt{5} \right )}{\sqrt{6}^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{3\left ( \sqrt{6}-\sqrt{5} \right )}{6-5}=\frac{3\left ( \sqrt{6}-\sqrt{5} \right )}{1}=3\left ( \sqrt{6}-\sqrt{5} \right )\\ \textrm{g}.\quad &\displaystyle \frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}=\displaystyle \frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\displaystyle \frac{\sqrt{3}\left ( \sqrt{6}+\sqrt{5} \right )}{\sqrt{6}^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{\sqrt{18}+\sqrt{15}}{6-5}=\frac{\sqrt{9.2}+\sqrt{15}}{1}=\left ( 3\sqrt{2}+\sqrt{15} \right )\\ \textrm{h}.\quad &\displaystyle \frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}=\displaystyle \frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}=\displaystyle \frac{\sqrt{3}\left ( \sqrt{6}-\sqrt{5} \right )}{\sqrt{6}^{2}-\sqrt{5}^{2}}\\ &=\displaystyle \frac{\sqrt{18}-\sqrt{15}}{6-5}=\frac{\sqrt{9.2}-\sqrt{15}}{1}=\left ( 3\sqrt{2}-\sqrt{15} \right )\\ \textrm{i}.\quad &\frac{\sqrt{3}}{\sqrt{6-2\sqrt{5}}}=\frac{\sqrt{3}}{\sqrt{5+1-2\sqrt{5.1}}}=\displaystyle \frac{\sqrt{3}}{\sqrt{5}-\sqrt{1}}=\frac{\sqrt{3}}{\sqrt{5}-1}\\ &=\frac{\sqrt{3}}{\sqrt{5}-1}\times \frac{\sqrt{5}+1}{\sqrt{5}+1}=\displaystyle \frac{\sqrt{3.5}+\sqrt{3.1}}{\sqrt{5}^{2}-1^{2}}=\frac{\sqrt{15}+\sqrt{3}}{5-1}\\ &=\displaystyle \frac{1}{4}\left ( \sqrt{15}+\sqrt{3} \right )\\ \textrm{j}.\quad &\frac{\sqrt{3}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{3}}{\sqrt{5+1+2\sqrt{5.1}}}=\displaystyle \frac{\sqrt{3}}{\sqrt{5}+\sqrt{1}}=\frac{\sqrt{3}}{\sqrt{5}+1}\\ &=\frac{\sqrt{3}}{\sqrt{5}+1}\times \frac{\sqrt{5}-1}{\sqrt{5}-1}=\displaystyle \frac{\sqrt{3.5}-\sqrt{3.1}}{\sqrt{5}^{2}-1^{2}}=\frac{\sqrt{15}-\sqrt{3}}{5-1}\\ &=\displaystyle \frac{1}{4}\left ( \sqrt{15}-\sqrt{3} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3&\textrm{Rasionalkan penyebut dan sederhanakanlah}\\ &\textrm{a}.\quad \displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}\\ &\textrm{b}.\quad\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}\\ &=\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}\times \displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{\sqrt{2}+\sqrt{5}-\sqrt{7}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{\left (\sqrt{2}+\sqrt{5} \right )^{2}-\left (\sqrt{7} \right )^{2}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{(2+2\sqrt{10}+5)-7}=\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{2\sqrt{10}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{2\sqrt{10}}\times \displaystyle \frac{\sqrt{10}}{\sqrt{10}}=\displaystyle \frac{\sqrt{20}+\sqrt{50}-\sqrt{70}}{2\times 10}\\ &=\displaystyle \frac{2\sqrt{5}+5\sqrt{2}+\sqrt{70}}{20} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\\ &=\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\times \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\left (\sqrt{2}+\sqrt{3} \right )^{2}-\left (\sqrt{5} \right )^{2}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{(2+2\sqrt{6}+3)-5}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{6}}\\ &=\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}\\ &=\displaystyle \frac{\sqrt{12}+\sqrt{18}+\sqrt{30}}{2\times 6}\\ &=\displaystyle \frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12} \end{aligned} \end{array}$

EKSPONEN (LANJUTAN 1)

 $\Large\textrm{C.2  Operasi Bilangan Bentuk Akar}$.

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

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

$\LARGE{ CONTOH SOAL}$.

$\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}\\\\ &\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}}}\\\\ &\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}$.

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

$\begin{array}{ll}\\ 4.&(\textbf{Matematika Dasar UM UGM 2008})\\ &\textrm{Bentuk sederhana dari}\\ &\qquad\qquad \displaystyle \frac{\sqrt[6]{x^{2}}\sqrt[3]{x^{2}\sqrt{x+1}}}{x\sqrt[6]{x+1}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sqrt[6]{x^{2}}\sqrt[3]{x^{2}\sqrt{x+1}}}{x\sqrt[6]{x+1}}\\ &=\displaystyle \frac{\sqrt[6]{x^{2}}.\sqrt[3.2]{\left (x^{2}.\sqrt{x+1} \right )^{2}}}{\sqrt[6]{x^{6}}.\sqrt[6]{x+1}}=\displaystyle \frac{\sqrt[6]{x^{2}}.\sqrt[6]{x^{4}.(x+1)}}{\sqrt[6]{x^{6}(x+1)}}\\ &=\displaystyle \frac{\sqrt[6]{x^{(2+4)}(x+1)}}{\sqrt[6]{x^{6}(x+1)}}=\displaystyle \frac{\sqrt[6]{x^{6}(x+1)}}{\sqrt[6]{x^{6}(x+1)}}\\ &=1 \end{aligned} \end{array}$.

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

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

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

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

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

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

DAFTAR PUSTAKA

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.

EKSPONEN

A. EKSPONEN

Eksponen (bilangan berpangkat) adalah bentuk perkalian berulang dari suatu bilangan dengan dirinya sendiri. Secara notasi 

Misalkan diketahui bahwa $a$ adalah suatu bilangan tidak nol dan $n$ adalah bilangan asli, maka bilangan ekponen atau bilangan berpangkat didefinisikan dengan:

$\LARGE a^{n}=\underset{n}{\underbrace{a\times a\times \times a\times ...\times a}}$

$\begin{aligned}\textrm{Bilangan}&:\\ a&\: \: \textrm{disebut basis atau bilangan pokok}\\ n&\: \: \textrm{disebut sebagai bilangan pangkat/eksponen} \end{aligned}$.

$\LARGE{ CONTOH SOAL}$.

$(1).\quad 3^{4}=3\times 3\times 3\times 3=81$

$(2).\quad 5^{4}=5\times 5\times 5\times 5=625$

$(3).\quad 2^{6}=2\times 2\times 2\times 2\times 2\times 2=64$

$(4).\quad 6^{7}=6\times 6\times 6\times 6\times 6\times 6\times 6=279936$

$(5).\quad (-3)^{3}=(-3)\times (-3)\times (-3)=-27$

$(6).\quad (-2)^{4}=(-2)\times (-2)\times (-2)\times (-2)=16$

$(7).\quad \left ( \displaystyle \frac{1}{5} \right )^{3}=\left ( \displaystyle \frac{1}{5} \right )\times \left ( \displaystyle \frac{1}{5} \right )\times \left ( \displaystyle \frac{1}{5} \right )= \displaystyle \frac{1}{125}$

$(8).\quad \left ( -\displaystyle \frac{1}{2} \right )^{3}=\left ( -\displaystyle \frac{1}{2} \right )\times \left (- \displaystyle \frac{1}{2} \right )\times \left ( -\displaystyle \frac{1}{2} \right )=- \displaystyle \frac{1}{8}$

$\Large\textrm{B. Sifat-Sifat Bilangan Pangkat Positif}$

$\begin{aligned}\\ 1.\quad&a^{m}.a^{n}=a^{m+n}\\ 2.\quad&a^{m}:a^{n}=a^{m-n}\\ 3.\quad&\left ( a^{m} \right )^{n}=a^{m.n},\: \: \textrm{syarat}\: \: a\neq 0\\ 4.\quad&\left ( ab \right )^{n}=a^{n}.b^{n}\\ 5.\quad&\left ( \frac{a}{b} \right )^{n}=\frac{a^{n}}{b^{n}},\: \: \textrm{syarat}\: \: b\neq 0 \end{aligned}$

Beberpa hal yang perlu diketahui juga, yaitu

$\begin{aligned}\\ 1.\quad&(a+b)^{2}=a^{2}+2ab+b^2\\ 2.\quad&(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\\ 3.\quad&\left ( a+\frac{1}{a} \right )^{2}=a^{2}+2+\displaystyle \frac{1}{a^{2}},\: \: \textrm{syarat}\: \: a\neq 0\\ \end{aligned}$

$\LARGE{ CONTOH SOAL}$

$\begin{aligned}\\ (1).\quad&2^{6} \times 2^{4} \times 2^{7} = 2^{6+4+7}=2^{17}\\ (2).\quad&2^{5} \times 3^{5} \times 7^{5} = \left ( 2 . 3 . 7 \right )^{5}=\left ( 42 \right )^{5}\\ (3).\quad&\displaystyle \frac{a^{3}.a^{7}.a^{6}}{a^{9}}=\displaystyle \frac{a^{3+7+6}}{a^{9}}=\frac{a^{16}}{a^{9}}=a^{16-9}=a^{7},\: \: \textrm{syarat}\: \: a\neq 0\\ \end{aligned}$

$\begin{aligned}(4).\quad\displaystyle \frac{3^{7}.7^{3}.2}{\left ( 42 \right )^{3}}&=\frac{2^{1}.3^{7}.7^{3}}{\left ( 2.3.7 \right )^{3}}=\frac{2^{1}.3^{7}.7^{3}}{2^{3}.3^{3}.7^{3}}\\ &=2^{1-3}.3^{7-3}.7^{3-3}=2^{-2}.3^{4}.7^{0}\\ &=\frac{1}{2^{2}}.3^{4}.1=\frac{3^{4}}{2^{2}} \end{aligned}$

$\begin{aligned}(5).\quad\displaystyle \frac{2^{2025}+2^{2026}+2^{2027}}{7}&=\displaystyle \frac{1.2^{2025}+2^{1}.2^{2025}+2^{2}.2^{2025}}{7}\\ &=\frac{\left ( 1+2+4 \right ).2^{2025}}{7}\\ &=\frac{7.2^{2025}}{7}\\ &=2^{2025} \end{aligned}$

$\begin{aligned}(6)\quad \displaystyle \frac{\left ( 2^{n+2} \right )^{2}-2^{2}.2^{2n}}{2^{n}.2^{n+2}}&=\displaystyle \frac{2^{2(n+2)}-2^{2}.2^{2n}}{2^{n}.2^{n}.2^{2}}\\ &=\displaystyle \frac{2^{2n}.2^{2.2}-2^{2}.2^{2n}}{2^{n+n}.2^{2}}\\ &=\displaystyle \frac{2^{2n}(2^{4}-2^{2})}{2^{2n}.2^{2}}\\ &=\displaystyle \frac{(2^{4}-2^{2})}{2^{2}}=\frac{16-4}{4}\\ &=\displaystyle \frac{12}{4}=3 \end{aligned}$

$\LARGE\textrm{C. Bentuk Akar}$

Bilangan bentuk akar di sini adalah kebalikan dari bilangan bentuk pangkat. Bilangan bentuk akar selanjutnya disebut bilangan irasional. Sebagai contoh $\sqrt{2}$, $\sqrt{3}$, $\sqrt{8}$, $\sqrt[3]{3}$, $\sqrt[3]{4}$, $\sqrt[3]{7}$ dan tapi ingat $\sqrt{4}$ dan  $\sqrt[3]{8}$ serta  $\sqrt[3]{27}$ adalah bukan bentuk akar, karena nantinya akan menghasilkan masing-masing 2 dan 3 serta 3.

$\begin{aligned}&\\ 1.\quad&a^{ \frac{1}{n}}=\sqrt[n]{a}\\ 2.\quad&a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\\ 3.\quad&a^{\frac{1}{2}}=\sqrt[2]{a^{1}}=\sqrt{a} \end{aligned}$.

$\textrm{Cara membaca}$.

$\begin{aligned}1.\quad&\sqrt[n]{p}\: \: \: \textbf{dibaca}\: \: \: \textrm{akar pangkat n dari p}\\ 2.\quad&\sqrt[n]{p^{2}}\: \: \: \textbf{dibaca}\: \: \: \textrm{akar pangkat n dari p kuadrat}\\ 3.\quad&\sqrt[n]{p^{3}}\: \: \: \textbf{dibaca}\: \: \: \textrm{akar pangkat n dari p pangkat tiga}\\ 4.\quad&\sqrt{p}\: \: \: \textbf{dibaca}\: \: \: \textrm{akar dari p}\: \: \: \textrm{atau}\\ &\qquad\qquad\qquad \textrm{akar kuadrat dari p}\\ &\qquad\qquad\qquad \textrm{ingat bahwa}:\: \: \sqrt{p}=\sqrt[2]{p} \end{aligned}$.

$\begin{aligned}\textrm{Defini}&\textrm{si}\\ \textrm{Jika}\: &\: a\: \: \textrm{dan}\: \: b\: \: \textrm{bilangan real dan}\\ &n\: \: \textrm{bilangan bulat positif, maka}:\\ &a^{n}=b\Leftrightarrow \sqrt[n]{b}=a\\ \textrm{keter}&\textrm{angan}:\\ \sqrt[n]{b}&\quad \textrm{disebut}\: \: \textbf{akar (radikal)}\\ b&\quad \textrm{disebut}\: \: \textbf{radikan}\\ &\quad \textrm{(bilangan pokok yang ditarik akarnya)}\\ n&\quad \textrm{disebut}\: \: \textbf{indeks}\\ &\quad (\textrm{pangkat akar}) \end{aligned}$.

$\Large\textrm{C.1  Bilangan Pangkat Pecahan}$.

Operasi Bilangan pangkat pecahan sama dengan operasi pangkat bilangan bulat.

$\LARGE{ CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&a^{.^{\frac{1}{2}}}\times a^{.^{\frac{1}{3}}}=a^{.^{\frac{1}{2}+\frac{1}{3}}}=a^{.^{\frac{5}{6}}}\\ 2.&a^{.^{\frac{1}{5}}}: a^{.^{\frac{1}{3}}}=a^{.^{\frac{1}{5}-\frac{1}{3}}}=a^{.^{-\frac{2}{15}}}\\ 3.&\left (a^{.^{\frac{2}{5}}} \right )^{\frac{4}{7}}=a^{.^{\frac{8}{35}}}\\ 4.&81^{.^{\frac{1}{2}}}=\left ( 9^{2} \right )^{.^{\frac{1}{2}}}=9^{1}=9\\ 5.&27^{.^{-\frac{2}{3}}}=\left ( 3^{3} \right )^{.^{-\frac{2}{3}}}=\left (3 \right )^{-2}=\displaystyle \frac{1}{3^{2}}=\frac{1}{9} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Sederhanakanlah bentuk berikut dan}\\ &\textrm{nyatakan hasilnya dalam pangkat positif}\\\\ &\textrm{a}.\quad \left ( 3p^{.^{\frac{5}{3}}}q^{.^{-\frac{3}{4}}} \right )\left ( 2p^{.^{-\frac{2}{3}}}q^{.^{\frac{5}{4}}} \right )\\\\ &\textrm{b}.\quad \displaystyle \frac{\left ( 8p^{.^{\frac{2}{3}}}q^{0}r^{.^{-\frac{1}{2}}} \right )}{\left ( 4p^{.^{-\frac{1}{2}}}q^{.^{-\frac{1}{3}}}r \right )}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left ( 3p^{.^{\frac{5}{3}}}q^{.^{-\frac{3}{4}}} \right )\left ( 2p^{.^{-\frac{2}{3}}}q^{.^{\frac{5}{4}}} \right )\\ &=3.2.p^{.^{\frac{5}{3}+\left ( -\frac{2}{3} \right )}}.q^{.^{-\frac{3}{4}+\frac{5}{4}}}\\ &=6.p^{.^{\frac{3}{3}}}q^{.^{\frac{2}{4}}}\\ &=6pq^{.^{\frac{1}{2}}} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\displaystyle \frac{\left ( 8p^{.^{\frac{2}{3}}}q^{0}r^{.^{-\frac{1}{2}}} \right )}{\left ( 4p^{.^{-\frac{1}{2}}}q^{.^{-\frac{1}{3}}}r \right )}\\ &=2.p^{.^{\frac{2}{3}-\left ( -\frac{1}{2} \right )}}q^{.^{0}-\left ( -\frac{1}{3} \right )}r^{.^{-\frac{1}{2}-1}}\\ &=2p^{.^{\frac{2}{3}+\frac{1}{2}}}q^{.^{\frac{1}{3}}}r^{.^{-\frac{3}{2}}}\\ &=2p^{.^{\frac{4+3}{6}}}q^{.^{\frac{1}{3}}}.r^{.^{-\frac{3}{2}}}\\ &=2p^{.^{\frac{7}{6}}}q^{.^{\frac{1}{3}}}.r^{.^{-\frac{3}{2}}}\\ &=\displaystyle \frac{2p^{.^{\frac{7}{6}}}q^{.^{\frac{1}{3}}}}{r^{.^{\frac{3}{2}}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Sederhanakanlah bentuk berikut dan}\\ &\textrm{nyatakan hasilnya dalam pangkat positif}\\ &\textrm{a}.\quad \left ( \displaystyle \frac{p^{3n+1}q^{n}}{p^{3n+4}q^{4n}} \right )^{\frac{1}{3}}\\ &\textrm{b}.\quad \left ( \displaystyle \frac{p^{-2}q^{3}}{p^{4}q^{-3}} \right )^{-\frac{1}{2}}\left ( \displaystyle \frac{p^{4}q^{-5}}{pq} \right )^{-\frac{1}{3}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left ( \displaystyle \frac{p^{3n+1}q^{n}}{p^{3n+4}q^{4n}} \right )^{\frac{1}{3}}\\ &=\left ( p^{(3n+1)-(3n+4)}q^{n-4n} \right )^{\frac{1}{3}}\\ &=\left ( p^{-3}q^{-3n} \right )^{\frac{1}{3}}\\ &=p^{-3.\frac{1}{3}}q^{-3n.\frac{1}{3}}\\ &=p^{-1}q^{-n}\\ &=\displaystyle \frac{1}{pq^{n}} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\left ( \displaystyle \frac{p^{-2}q^{3}}{p^{4}q^{-3}} \right )^{-\frac{1}{2}}\left ( \displaystyle \frac{p^{4}q^{-5}}{pq} \right )^{-\frac{1}{3}}\\ &=\left ( \displaystyle \frac{p^{-2.(-\frac{1}{2})}q^{3.(-\frac{1}{2})}}{p^{4.(-\frac{1}{2})}q^{-3.(-\frac{1}{2})}} \right )\left ( \displaystyle \frac{p^{4.(-\frac{1}{3})}q^{-5.(-\frac{1}{3})}}{p^{.^{-\frac{1}{3}}}q^{.^{-\frac{1}{3}}}} \right )\\ &=\displaystyle \frac{p^{1}q^{.^{-\frac{3}{2}}}}{p^{-2}q^{.^{\frac{3}{2}}}}\times \frac{p^{.^{-\frac{4}{3}}}q^{.^{\frac{5}{3}}}}{p^{.^{-\frac{1}{3}}}q^{.^{-\frac{1}{3}}}}\\ &=p^{1-(-2)+(-\frac{4}{3})-(-\frac{1}{3})}q^{-\frac{3}{2}-\frac{3}{2}+\frac{5}{3}-(-\frac{1}{3})}\\ &=p^{3-\frac{3}{3}}q^{-\frac{6}{2}+\frac{6}{3}}\\ &=p^{3-1}q^{-3+2}\\ &=p^{2}q^{-1}\\ &=\displaystyle \frac{p^{2}}{q} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Jabarkanlah bentuk}\\ &\qquad\qquad\quad \left ( 2m^{.^{\frac{3}{2}}}+n^{.^{\frac{3}{4}}} \right )^{2}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\left ( 2m^{.^{\frac{3}{2}}}+n^{.^{\frac{3}{4}}} \right )^{2}\\ &=\left ( 2m^{.^{\frac{3}{2}}} \right )^{2}+2\left ( 2m^{.^{\frac{3}{2}}} \right )\left ( n^{.^{\frac{3}{4}}} \right )+\left ( n^{.^{\frac{3}{4}}} \right )^{2}\\ &\textrm{INGAT}\: :\: \: \color{black}\left ( A+B \right )^{2}=A^{2}+2AB+B^{2}\\ &=2^{2}m^{.^{\frac{3.2}{2}}}+2.2.m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{4}}}+n^{.^{\frac{3.2}{4}}}\\ &=4m^{3}+4m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{4}}}+n^{.^{\frac{3}{2}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jabarkanlah bentuk}\\ &\qquad\qquad\quad \left ( 2m^{.^{\frac{3}{2}}}-n^{.^{\frac{3}{4}}} \right )^{3}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\left ( 2m^{.^{\frac{3}{2}}}-n^{.^{\frac{3}{4}}} \right )^{3}\\ &=\left ( 2m^{.^{\frac{3}{2}}} \right )^{3}-3\left ( 2m^{.^{\frac{3}{2}}} \right )^{2}\left ( n^{.^{\frac{3}{4}}} \right )+3\left ( 2m^{.^{\frac{3}{2}}} \right )\left ( n^{.^{\frac{3}{4}}} \right )^{2}-\left ( n^{.^{\frac{3}{4}}} \right )^{3}\\ &\textrm{INGAT}\: :\: \: \color{black}\left ( A-B \right )^{3}=A^{3}-3A^{2}B+3AB^{2}-B^{3}\\ &=2^{3}m^{.^{\frac{3.3}{2}}}-3.2^{2}.m^{.^{\frac{3.2}{2}}}n^{.^{\frac{3}{4}}}+3.2.m^{.^{\frac{3}{2}}}n^{.^{\frac{3.2}{4}}}-n^{.^{\frac{3.3}{4}}}\\ &=8m^{.^{\frac{9}{2}}}-12m^{3}n^{.^{\frac{3}{4}}}+6m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{2}}}-n^{.^{\frac{9}{2}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Jabarkanlah bentuk berikut}\\\\ &\textrm{a}.\quad \left ( 2p^{.^{\frac{1}{2}}}-3q^{.^{\frac{1}{2}}} \right )\left ( p^{.^{\frac{1}{2}}}+4q^{.^{\frac{1}{2}}} \right )\\\\ &\textrm{b}.\quad \left ( p^{.^{\frac{1}{3}}}-q^{.^{\frac{1}{3}}} \right )\left ( p^{.^{\frac{2}{3}}}+p^{.^{\frac{1}{3}}}q^{.^{\frac{1}{3}}}+q^{.^{\frac{2}{3}}} \right )\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left ( 2p^{.^{\frac{1}{2}}}-3q^{.^{\frac{1}{2}}} \right )\left ( p^{.^{\frac{1}{2}}}+4q^{.^{\frac{1}{2}}} \right )\\ &=2\left (p^{.^{\frac{1}{2}}} \right )^{2}+2.4.p^{.^{\frac{1}{2}}}q^{.^{\frac{1}{2}}}-3q^{.^{\frac{1}{2}}}.p^{.^{\frac{1}{2}}}-3.4.\left (q^{.^{\frac{1}{2}}} \right )^{2}\\ &=2p^{1}+8q^{.^{\frac{1}{2}}}q^{.^{\frac{1}{2}}}-3p^{.^{\frac{1}{2}}}q^{.^{\frac{1}{2}}}-12.q^{1}\\ &=2p+5(pq)^{.^{\frac{1}{2}}}-12q \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\left ( p^{.^{\frac{1}{3}}}-q^{.^{\frac{1}{3}}} \right )\left ( p^{.^{\frac{2}{3}}}+p^{.^{\frac{1}{3}}}q^{.^{\frac{1}{3}}}+q^{.^{\frac{2}{3}}} \right )\\ &=p^{.^{\frac{1+2}{3}}}+\left (p^{.^{\frac{1}{3}}} \right )^{2}q^{.^{\frac{1}{3}}}+p^{.^{\frac{1}{3}}}q^{.^{\frac{2}{3}}}-p^{.^{\frac{2}{3}}}q^{.^{\frac{1}{3}}}-p^{.^{\frac{1}{3}}}\left (q^{.^{\frac{1}{3}}} \right )^{2}-q^{.^{\frac{1+2}{3}}}\\ &=p^{1}+0+0-q^{1}\\ &=p-q \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. 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.

Fungsi Kuadrat

 $\begin{array}{|l|l|}\hline \textrm{Pengertian}&\begin{aligned}&\textrm{Suatu fungsi yang berbentuk}\\ &f(x)=ax^{2}+bx+c\\ & a,\: b,\: c,\: \in \mathbb{R},\: a\neq 0 \end{aligned}\\\hline \textrm{Grafik Fungsi}&\textrm{Keterangan}\\\hline \textrm{Titik potong sumbu x}&\textrm{Jika ada}\\\hline &\begin{aligned}&\textrm{untuk titik potong}\\ &\textrm{terhadap sumbu x }\\ &\textrm{Jika y = 0 maka }\\ &ax^{2}+bx+c=0\\ &\textrm{Selanjutnya tinggal}\\ &\textrm{menentukan nilai D}\\ &D=b^{2}-4ac\: \: \textrm{adalah}\\ &\: \: \: \: \: \: \: \: \: \textrm{nilai diskriminan}.\\ &\textrm{Jika} \: D>0\\ &\textrm{maka grafik}\\ &\textrm{memotong sumbu x}\\ &\textrm{di dua tempat berbeda}\\ &\textrm{yaitu di} \: (x_{1},0)\: \textrm{dan}\: (x_{2},0).\\ &\textrm{dan jika D = 0}\\ &\textrm{maka grafik}\\ &\textrm{ hanya menyinggung}\\ &\textrm{sumbu x di satu titik}\\ &\textrm{yaitu di }\: (x_{1},0)\\ &\textrm{dan jika}\: D<0 \\ &\textrm{maka grafik}\\ &\textrm{tidak memotong}\\ &\textrm{atau menyinggung sumbu x} \end{aligned}\\\hline \textrm{Titik potong sumbu y}&\begin{aligned}&\textrm{titik potong terhadap}\\ &\textrm{sumbu y, jika x = 0}\\ &y=f(x)=ax^{2}+bx+c\\ &y=f(0)=a(0)^{2}+b(0)+c\\ &y=c \end{aligned}\\\hline \textrm{Sumbu Simetri (SS)}&x=\displaystyle \frac{-b}{2a}\\\hline \textrm{Titik Puncak}&\left ( \displaystyle \frac{-b}{2a},\displaystyle \frac{D}{-4a} \right )\\\hline \textrm{Posisi grafik}&\textrm{Jika}\: a>0\: \textrm{maka}\\ &\textrm{grafik terbuka ke atas}\\ &\textrm{Dan jika nilai}\: a<0\: \textrm{maka}\\ &\textrm{grafik terbuka ke bawah}\\\hline \end{array}$.

Selanjutnya cara membuat grafik fungsi kudratnya adalah sebagai berikut:

$\begin{array}{|c|c|}\hline \textrm{Jika memotong sumbu}-\textrm{X}&\textrm{Jika menyinggung sumbu}-\textrm{X}\\ \textrm{di titik}\: \left ( x_{1},0 \right )\: \textrm{dan}\: \left ( x_{2},0 \right )&\textrm{di titik}\: \left ( x_{1},0 \right )\: \textrm{dan melalui}\\ \textrm{dan melalui sebuah titik lain}&\textrm{sebuah titik lain} \\\hline &\\ y=f(x)=a\left ( x-x_{1} \right )\left ( x-x_{2} \right )&y=f(x)=a\left ( x-x_{1} \right )^{2}\\ &\\\hline \textrm{Jika grafik fungsi itu melalui}&\textrm{Jika grafik fungsi itu melalui}\\\hline \textrm{Titik puncak}\: \: P\left ( x_{p},y_{p} \right )\: \textrm{dan}&\textrm{tiga buah titik yaitu}\: \left ( x_{1},y_{1} \right )\\ \textrm{sebuah titik lain}&\left ( x_{2},y_{2} \right )\: \: \textrm{dan}\: \: \left ( x_{3},y_{3} \right )\\\hline &\\ y=f(x)=a\left ( x-x_{p} \right )^{2}+y_{p}&y=f(x)=ax^{2}+bx+c\\ &\\\hline \end{array}$.

$\LARGE{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: f\: \: \textrm{adalah fungsi linear dengan}\\ & f(2)-f(-2)=8,\\ & \textrm{maka nilai dari}\: \: f(4)-f(-2)\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\\ &f(x)=ax+b\\ &f(2)-f(-2)\\ &=\left (a(2)+b \right )-\left ( a(-2)+b \right )=8\\ &8=2a+2a\\ &8=4a\\ &2=a\\ &f(x)=2x+b,\quad \textrm{dengan}\: \: b\: \: \textrm{konstan}\\ &\textrm{Sehingga nilai}\quad\\ &f(4)-f(-2)=\left (2(4)+b \right )-\left (2(-2)+b \right )\\ &=8+b+4-b\\ &=12 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Ubahlah}\: \: 8-6x-x^{2}\: \: \textrm{ke dalam bentuk}\\ & a-(x+b)^{2},\: \textrm{selanjutnya tentukan}\\ & \textrm{daerah hasil dari}\: \: f(x)=8-6x-x^{2}\\ & \textrm{untuk}\: \: x\: \: \textrm{bilangan real}\\ &\qquad(\textit{NTU Entrance Examination AO-level})\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|l|}\hline 1.&\textrm{Diketahui}\\\hline &\begin{aligned}\textrm{Misal}\quad\qquad&\\ 8-6x-x^{2}&=f(x)\\ f(x)&=-x^{2}-6x+8\\ &=-\left ( x^{2}+6x-8 \right )\\ &=-\left ( x^{2}+6x+9-17 \right )\\ &=-\left ( (x+3)^{2}-17 \right )\\ &=-(x+3)^{2}+17\\ & \end{aligned}\\\hline 2.&\textrm{Mencari koordinat}\: \: \left ( x_{SS},y_{SS} \right )\\\hline &\begin{aligned}f(x)&=-x^{2}-6x+8\left\{\begin{matrix} a=-1\\ b=-6\\ c=\: \: 8\: \: \end{matrix}\right.\\ \textrm{Maka}&\\ x_{SS}&=\frac{-b}{2a}=\displaystyle \frac{-(-6)}{2(-1)}\\ &=-3\\ y_{SS}&=f(-3)=-\left ( -3+3 \right )^{2}+17=17\\ \therefore &\left ( x_{SS},y_{SS} \right )=(-3,17) \end{aligned}\\\hline 3.&\textrm{Nilai fungsi}\\\hline &\begin{aligned}\textrm{Karena}&\: \: a=-1<0\\ \textrm{maka f}&\textrm{ungsi menghadap}\\ \textbf{ke ba}&\textbf{wah},\: \: \textrm{sehingga}\\ \textrm{daerah}&\: \: \textrm{hasilnya}\: \: \left (R_{f} \right )\\ \textrm{adalah}&:\\ &\left \{ -\infty <y\leq 17 \right \}\\ &\\ &\textrm{Berikut ilustrasinya} \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah akar-akar dari }\\ &\textrm{persamaan kuadrat}\: \: x^{2}+mx+m=0,\\ &\textrm{maka nilai}\: \: m\: \: \textrm{yang menyebabkan }\\ &\textrm{jumlah kuadrat akar-akar mencapai}\\ &\textrm{minimum adalah}\: ....\\ &\qquad \: \textbf{(UM UNDIP 2014 Mat Das)}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \: x^{2}+mx+m=0\\ & \textbf{persamaan kuadrat}\: \textrm{dalam}\: \: x,\\ & \textrm{maka}\\ &x^{2}+mx+m=x^{2}-(\alpha +\beta )x+(\alpha \beta )=0\\ &\begin{cases} \alpha +\beta &=-m \\ & \\ \alpha \beta &=m \end{cases}\\ &\textrm{Selanjutnya}\\ &\alpha ^{2}+\beta ^{2}=\left ( \alpha +\beta \right )^{2}-2\alpha \beta\\ &=(-m)^{2}-2m\: \: \textrm{dan dapat kita tuliskan sebagai}\\ &f(m)=m^{2}-2m\begin{cases} a &=1 \\ b &=-2 \\ c &=0 \end{cases} \\ &\textrm{fungsi kuadrat dalam}\: \: m,\\ &\textrm{sehingga kita perlu mencari titik}\: \: \left ( m_{SS},f\left ( m_{SS} \right ) \right ),\\ & \textrm{tetapi yang kita perlukan}\\ &\textrm{cuma}\: \: m-\textrm{nya saja, yaitu}:\: \: m=m_{SS},\\ &\textrm{dengan}\quad m_{SS}=\displaystyle \frac{-b}{2a}=\frac{-(-2)}{2.1}=1 \end{aligned} \end{array}$.

Link Materi

https://ahmadthohir1098.blogspot.com/2024/11/kumpulan-materi-matematika-ma-sma-kelas.html

Sistem Pertidaksamaan Linear

 B. Sistem Pertidaksamaan Linear

$\begin{array}{ll}\\ &\textbf{BENTUK UMUM}\\ &\begin{cases} ax+by<c \\ ax+by\leq c \\ ax+by>c \\ ax+by\geq c \end{cases}\\\\ &\textbf{LANGKAH-LANGKAH}\\ &\textrm{dalam membuat gambar grafik persamaan linear}\\ &\: \textrm{adalah sebagai berikut}:\\ &\bullet\quad \textrm{membuat gambar grafik}\: \: ax+by=c\\ &\quad \: \: \textrm{untuk batas wilayahnya}\\ &\bullet \quad \textrm{menyelidiki wilayah yang dimaksud di sekitar}\\ &\quad \: \: \textrm{garis} \: \: ax+by=c\\ &\bullet \quad \textrm{ambillah sebuah titik}\: \left ( x_{0},y_{0} \right )\: \textrm{sembarang}\\ &\: \: \quad \textrm{kemudian substitusikan ke pertidaksamaan}\\ &\quad \: \: ax+by\: ....\: c\\ &\bullet \quad \textrm{jika diperoleh nilai ketaksamaan yang benar},\\ &\: \: \quad \textrm{maka daerah di mana titik uji}\: \left ( x_{0},y_{0} \right )\\ &\: \: \quad \textrm{berada merupakan wilayah penyelesaiannya}\\ &\: \: \quad \textrm{demikian juga sebaliknya} \end{array}$

$\LARGE\fbox{CONTOH SOAL}$

$\begin{array}{l}\\ 1.&\textrm{Gambarlah himpunan penyelesaian (HP)}\\ &\textrm{dari pertidaksamaan linear berikut}\\ &\textrm{a}.\quad 3x+2y< 6\\ &\textrm{b}.\quad 3x+2y\leq 6\\ &\textrm{c}.\quad 3x+2y> 6\\ &\textrm{d}.\quad 3x+2y\geq 6\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Mula}-\textrm{mula kita gambar garis}\: \: 3x+2y=6\\\\ &\begin{array}{|c|c|c|}\hline \textrm{Komponen}&\textrm{pada}&\textrm{pada}\\ \textrm{titik}&\textrm{sumbu}-y&\textrm{sumbu}-x\\\hline x&0&2\\\hline y&3&0\\\hline (x,y)&(0,3)&(2,0)\\\hline \end{array}\\\\ &\textrm{Selanjutnya gambar grafiknya sebagai berikut}. \end{aligned} \end{array}$

Dan berikut untuk wilayah dan juga batas-batas untuk pertidalsamaan

$3x+2y<6$

Kita dapat menggunakan titik uji untuk memastikan kondisi gambar di atas, yaitu di antaranya

$\begin{array}{|c|c|c|}\hline \textrm{Titik}&\textrm{Pengujian}&\textrm{Keterangan}\\ &\textrm{Uji}&3x+2y<6\\\hline (0,0)&3(0)+2(0)=0<\textbf{6}&\textrm{Dalam wilayah}\\\hline (0,1)&3(0)+2(1)=2<\textbf{6}&\textrm{Dalam wilayah}\\\hline (1,0)&3(1)+2(0)=3<\textbf{6}&\textrm{Dalam wilayah}\\\hline (1,1)&3(1)+2(1)=5<\textbf{6}&\textrm{Dalam wilayah}\\\hline (0,2)&3(0)+2(2)=4<\textbf{6}&\textrm{Dalam wilayah}\\\hline (2,0)&3(2)+2(0)=6=\textbf{6}&\textrm{Di luar wilayah}\\\hline (2,2)&3(2)+2(2)=10>\textbf{6}&\textrm{Di luar wilayah}\\\hline (0,3)&3(0)+2(3)=6=\textbf{6}&\textrm{Di luar wilayah}\\\hline (3,0)&3(3)+2(0)=9>\textbf{6}&\textrm{Di luar wilayah}\\\hline (3,3)&3(3)+2(3)=15>\textbf{6}&\textrm{Di luar wilayah}\\\hline \vdots &\vdots&\vdots \\\hline \end{array}$

Dan berikut untuk wilayah yang memenuhi  $"3x+2y\leq 6$

$\begin{array}{ll}\\ 2.&\textrm{Selesaikanlah pertidaksamaan berikut}\\ &\textrm{a}.\quad 12x+2>4x+6\\ &\textrm{b}.\quad 2-3x<6-x\\ &\textrm{c}.\quad 6x+1\geq 2\\ &\textrm{d}.\quad \displaystyle \frac{2-3x}{2}<\frac{3-x}{3}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{a}.\: \: 12x&+2>4x+6\\ 12x&-4x>6-2\\ 8x&>4\\ x&>\displaystyle \frac{1}{2} \end{aligned}\\ &\begin{aligned}\color{black}\textrm{b}.\: \: &2-3x<6-x\\ &-3x+x<6-2\\ &-2x<4\: \: \textrm{dikali}\: \: (-1)\\ &2x>-4\: \: (\textrm{tanda berubah})\\ &x>-2 \end{aligned}\\ &\begin{aligned}\textrm{c}.\: \: 6x&+1\geq 2\\ 6x&\geq 2-1\\ x&\geq \displaystyle \frac{1}{6} \end{aligned}\\ &\begin{aligned}\textrm{d}.\: \: \: \: \displaystyle \frac{2-3x}{2}&<\frac{3-x}{3}\\ 3(2-3x)&<2(3-x)\\ 6-9x&<6-2x\\ -9x+2x&<6-6\\ -7x&<0\: \: \textrm{di kali}\: \: (-1)\\ 7x&>0\: \: (\textrm{tanda berubah})\\ x&>\displaystyle \frac{0}{7}\\ x&>0 \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Heryadi, D. 2007. Modul Matematikauntuk SMK Kelas X. Bogor: YUDHISTIRA.
2. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo. PT. TIGA SERANGKAI PUSTAKA MANDIRI.

Sistem Persamaan Linear

 A. Sistem Persamaan Linear

Sistem persamaan linear adalah adalah kumpulan dari beberapa persamaan linear di mana koefisien-koefisien persamaannya berupa bilangan real dan anatar variabel saling ada keterkaitan

$\textrm{1. Sistem Persamaan Linear Dua Variabel}$

Sistem Persamaan Linear Dua Variabel yang selanjutnya disingkat dengan SPLDV memiliki bentuk umum sebagai berikut:

$\begin{aligned}&\left\{\begin{matrix} a_{1}x+b_{1}y=c_{1}\\ a_{2}x+b_{2}y=c_{2} \end{matrix}\right. \end{aligned}$

Keterangan:

• $x,y\: \: \textrm{adalah variabel}$.
• $a_{1},a_{2}\: \: \textrm{koefisien}\: \: x$
• $b_{1},b_{2}\: \: \textrm{koefisien}\: \: y$.
• $c_{1},c_{2}\: \: \textrm{adalah konstanta}$.
• $a_{1},a_{2},b_{1},b_{2},c_{1},\: \: \textrm{dan}\: \: c_{2}\: \: \textrm{adalah bilangan riil}$.

$\textrm{2. Sistem Persamaan Linear Tiga Variabel}$

$\begin{array}{l}\\ \underline{\textbf{Bentuk Umum}}&:\\ &\begin{cases} a_{1}x+b_{1}y+c_{1}z=d_{1} \\ a_{2}x+b_{2}y+c_{2}z=d_{2} \\ a_{3}x+b_{3}y+c_{3}z=d_{3} \end{cases}\\\\ \qquad \quad \textbf{Keterangan}&\bullet \quad a_{1},\: a_{2},\: a_{3},\\ &\, \: \: \quad b_{1},\: b_{2},\: b_{3},\\ &\, \: \: \quad c_{1},\: c_{2},\: c_{3},\\ &\, \: \: \quad d_{1},\: d_{2},\: d_{3}\\ &\: \: \quad \textrm{semuanya adalah bilangan real} \end{array}$

$\textrm{B. Penyelesaian Sistem persamaan Linear}$

Menentukan penyelesaian atau himpunan penyelesaian (HP) dari sistem persamaan linear baik yang terdiri dari dua variabel ataupun tiga variabel adalah menentukan pasangan koordinat yang memenuhi sistem persamaan tersebut di bilangan riil. Adapun cara menyelesaikan sistem persamaan linear ini

• Metode Substitusi
• Metode Eliminasi
• Metode Eliminasi-Substitusi 
• Metode Determinan Matrik
• Metode Invers Matrik (Matrik Persegi minimal ordo 2x2)

$\textrm{1. Metode Eliminasi-Substitusi}$

Adapun langkah-langkah dalam penyelesaian model tipe ini (Metode Substitusi dan Metode Eliminasi mengikuti karena prosesnya terangkum di langkah gabungan ini) adalah:

• buatlah dua buah kelompok persamaan yang memungkinkan dapat disederhanakan (kalau bisa ambil yang termudah dan sederhana menurut Anda)
• Salah satu variabel dihilangkan dengan cara menyamakan koefisien variabel yang bersangkutan kemudian mengeliminasikan dengan persamaan linear yang dipilih pada langkat pertama tadi.
• Nilai variabel yang didapatkan disubstitusikan ke dalam salah satu persamaan pada langkah pertama tadi juga.
• Jika diperlukan lagi, prinsipnya kembali pada poin pertama tadi

$\LARGE\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah penyelesaian SPLDV dari}\\ &\begin{cases} 2x-y=7 \\ x-y=-1 \end{cases}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Mis}&\textrm{alkan}\\ &\begin{cases} 2x-y=7&.....(1) \\ x-y=-1&.....(2) \end{cases}\\ &\begin{aligned}&\textrm{dari persamaan}\: (2)\: \textrm{didapatkan}\\ &x=y-1.\: \textrm{Bentuk ini kemudian}\\ &\textrm{kita substitusikan ke}\\ &\textrm{persamaan}\: \: (1). \end{aligned}\\ &\begin{aligned}2x-y&=7\\ 2\left ( y-1 \right )-y&=7\\ 2y-2-y&=7\\ y&=9\quad .....(3)\\ \textrm{Selanjutnya}&\: \textrm{nilainya kita}\\ \textrm{substitusikan ke}&\: \textrm{persamaan}\: \: (2)\\ x&=y-1\\ x&=9-1\\ x&=8 \end{aligned} \\ &\begin{aligned}&\textrm{Sehingga},\\ &\begin{cases} x =8 \\ y =9 \end{cases}\\ &\textrm{Jadi, HP}=\left \{ (8,9) \right \} \end{aligned}\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah penyelesaian SPLDV dari}\\ &\begin{cases} 2x-y+z=-4 \\ 2x-y-2z=-3\\ x+3y-z=0 \end{cases}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Per}&\textrm{hatikan misal}\\ &\begin{cases} 2x-y+z=-4..........(1) \\ 2x-y-2z=-3........(2)\\ x+3y-z=0.............(3) \end{cases}\\ &\begin{aligned}&\textrm{dari persamaan}\: (2)\: \textrm{didapatkan}\\ &2x-y=2z-3.\: \textrm{Bentuk ini}\\ &\textrm{kita substitusikan ke}\\ &\textrm{persamaan}\: \: (1). \end{aligned} \\ &\begin{aligned}2x-y+z&=-4\\ (2z-3)+z&=-4\\ 3z&=-1\\ z&=-\frac{1}{3}\quad .....(4)\\ \textrm{Selanjutnya}&\: \textrm{nilai tersebut kita}\\ \textrm{substitusikan ke}&\: \textrm{pers.}\: \: (2)\: \textrm{dan}\: (3)\\ \end{aligned}\\ &\begin{aligned}&\textrm{Selanjutnya}\\ &\begin{cases} 6x-3y=-11.....(2) \\ x+3y=-\frac{1}{3}.....(3) \end{cases} \end{aligned}\\ &\begin{aligned}&\textrm{dengan cara seperti}\\ &\textrm{poin 1.a kita akan}\\ &\textrm{mendapatkan nilai}\\ x&=-\frac{34}{21}\: \textrm{dan}\\ y&=\frac{3}{7} \end{aligned} \\ &\begin{aligned}&\textrm{Sehingga},\\ &\begin{cases} x =-\frac{34}{21} \\ y =\frac{3}{7}\\ z=-\frac{1}{3} \end{cases}\\ &\textrm{HP}=\left \{ \left ( -\frac{34}{21},\frac{3}{7},-\frac{1}{3} \right ) \right \} \end{aligned} \end{aligned} \end{array}$

TAMBAHAN MATERI

$\textrm{2. Metode determinan Matriks}$

Perhatikan kemabil bentuk SPLDV dan SPLTV berikut:

$\begin{cases} a_{1}x+b_{1}y=c_{1} \\ a_{2}x+b_{2}y=c_{2} \end{cases}$ 

dan

$\begin{cases} a_{1}x+b_{1}y+c_{1}z=d_{1}\\ a_{1}x+b_{1}y+c_{1}z=d_{1} \\ a_{1}x+b_{1}y+c_{1}z=d_{1} \end{cases}$

Metode determinat matriks adalah penyelesaian nilai tidap variabel dengan menggunakan determinan berikut:

Misalkan saja diberikan:

$\begin{aligned}&\begin{aligned}ax+by&=p\\ cx+dy&=q \end{aligned}\\\\ &\textrm{dan}\\\\ &\begin{aligned}ax+by+cz&=r\\ dx+ey+fz&=s\\ gx+hy+iz&=t \end{aligned}\\ \end{aligned}$

maka penyelesaian dengan model matriks adalah:

$\begin{array}{|c|c|c|}\hline \textrm{Metode}&\textbf{SPLDV}&\textbf{SPLTV}\\\hline \textrm{Determinan}&\begin{aligned}x&=\displaystyle \frac{\begin{vmatrix} p & b\\ q & d \end{vmatrix}}{\begin{vmatrix} a & b\\ c & d \end{vmatrix}}\\ &\textrm{dan}\\ y&=\displaystyle \frac{\begin{vmatrix} a & p\\ c & q \end{vmatrix}}{\begin{vmatrix} a & b\\ c & d \end{vmatrix}}\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}x&=\displaystyle \frac{\begin{vmatrix} r & b & c\\ s & e & f\\ t & h & i \end{vmatrix}}{\begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}}\\ &\textrm{dan}\\ y&=\displaystyle \frac{\begin{vmatrix} a & r & c\\ d & s & f\\ g & t & i \end{vmatrix}}{\begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}}\\ &\textrm{serta}\\ z&=\displaystyle \frac{\begin{vmatrix} a & b & r\\ d & e & s\\ g & h & t \end{vmatrix}}{\begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}} \end{aligned}\\\hline \end{array}$

Sebagai catatan:

$\begin{aligned}&\begin{vmatrix} a & b\\ c & d \end{vmatrix}=ad-bc\\ &\textrm{dan}\\ &\begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}=a\begin{vmatrix} e & f\\ h & i \end{vmatrix}-b\begin{vmatrix} d & f\\ g & i \end{vmatrix}+c\begin{vmatrix} d & e\\ g & h \end{vmatrix} \end{aligned}$

$\LARGE\fbox{CONTOH SOAL}$

Mari kita buka lagi contoh sebelumnya dengan soal yang sama di SINI

dan kearang penyelesaian dari soal tersebut akan diselesaikan dengan cara determinan matriks (cara Cramer sesuai nama penemunya) berikut:

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah dengan metode matriks}\\ &\textrm{(cara Cramer) SPLDV berikut}:\\ &\begin{cases} 2x-y & =7 \\ x-y & =-1 \end{cases}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}x&=\displaystyle \frac{\begin{vmatrix} 7 & -1\\ -1 & -1 \end{vmatrix}}{\begin{vmatrix} 2 & -1\\ 1 & -1 \end{vmatrix}}=\frac{7(-1)-(-1).(-1)}{2.(-1)-(-1).1}\\ &=\displaystyle \frac{-7-1}{-2+1}=\frac{-8}{-1}=8\\ y&=\displaystyle \frac{\begin{vmatrix} 2 & 7\\ 1 & -1 \end{vmatrix}}{\begin{vmatrix} 2 & -1\\ 1 & -1 \end{vmatrix}}=\frac{2(-1)-(7).1}{2.(-1)-(-1).1}\\ &=\displaystyle \frac{-2-7}{-2+1}=\frac{-9}{-1}=9\\ \textrm{J}&\textrm{adi}\: \: (x,y)=(8,9) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah dengan metode matriks}\\ &\textrm{(cara Cramer) SPLTV berikut}:\\ &\begin{cases} 2x-y+z & =-4 \\ 2x-y-2z & =-3\\ x+3y-z&=0 \end{cases}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}x&=\displaystyle \frac{\begin{vmatrix} -4 & -1&1\\ -3 & -1&-2\\ 0&3&-1 \end{vmatrix}}{\begin{vmatrix} 2 & -1&1\\ 2 & -1&-2\\ 1&3&-1 \end{vmatrix}}\\ x&=\displaystyle \frac{-4\begin{vmatrix} -1&-2\\ 3&-1 \end{vmatrix}+1\begin{vmatrix} -3 & -2\\ 0 & -1 \end{vmatrix}+1\begin{vmatrix} -3 & -1\\ 0 & 3 \end{vmatrix}}{2\begin{vmatrix} -1 & -2\\ 3 & -1 \end{vmatrix}+1\begin{vmatrix} 2 & -2\\ 1 & -1 \end{vmatrix}+1\begin{vmatrix} 2 & -1\\ 1 & 3 \end{vmatrix}} \\ &=\displaystyle \frac{-4(1+6)+1(3-0)+1(-9-0)}{2(1+6)+1(-2+2)+1(6+1)}\\ &=\displaystyle \frac{-28+3-9}{14+0+7}\\ &=\frac{-34}{21}\\ y&=.... \\ z&=....\\ \textrm{J}&\textrm{adi}\: \: (x,y,y)=\left ( -\displaystyle \frac{34}{21},\frac{3}{7},-\frac{1}{3} \right ) \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Johanes, Kastola & Sulasim. 2006. Kompetensi Matematika 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA

Materi Link

https://ahmadthohir1098.blogspot.com/2024/11/kumpulan-materi-matematika-ma-sma-kelas.html

Materi Persiapan TKA SMA-MA 2025

Materi Transformasi Fungsi

KELAS XII - KUMPULAN MATERI MATEMATIKA PEMINATAN MA-SMA (KURTILAS REVISI)

 A. Kelas X (Sepuluh)

A. 1 Fungsi Eksponensial dan Fungsi Logaritma

• materi eksponen 1, materi fungsi eksponen 2, lanjutan fungsi eksponen, lanjutan materi 2 fungsi ekponen, lanjutan materi 3 fungsi eksponen, Persamaan eksponen.
• contoh soal eksponen 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6 , contoh 7 , contoh 8 , contoh 9 , contoh 10 
• materi logaritma 
• materi logaritma lanjutan
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6
• materi logaritma lanjutan 2
• materi persamaan logaritma 1, persamaan 2, persamaan 3, persamaan 4, persamaan 5. aplikasi logaritma, pengayaan (operasi logaritma natural)
• contoh soal 7 , contoh 8 , contoh 9 , contoh 10, contoh 11, contoh 12, contoh 13.

A. 2 Vektor

• materi vektor, operasi vektor, vektor di dimensi dua, perkalian skalar dua vektor, proyeksi ortogonal suatu vektor, vektor dimensi tiga, lanjutan vektor dimensi tiga, operasi vektor berdimensi tiga, lanjutan 1 operasi vektor berdimensi tiga, lanjutan 2 operasi vektor berdimensi tiga, proyeksi orthogonal suatu vektor di dimensi tiga, 
• contoh soal 1, contoh 2, contoh 3, contoh 4, (urutan) contoh soal 5, contoh 6, contoh 7, contoh 8, contoh 9, contoh 10, contoh 11, contoh 12, contoh 13.

B. Kelas XI (Sebelas)

B. 1 Persamaan Trigonometri

• materi persamaan trigonometri, identitas trigonometri, menentukan nilai perbandingan trigonometri pada segitiga siku-siku, relasi sudut, persamaan trigonometri sederhana.
• lanjutan materi 1 , materi 2 , materi 3
• materi baru
• persamaan tigonometri: identitas trigonometri, menentukan nilai perbandingan trigonometri, sudut-sudut berelasi, persamaan trigonometri sederhana, persamaan bentuk a sin x + b cos x=c, grafik fungsi trigonometri, 
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6
• soal tengah semester gasal : contoh 1, contoh 2, contoh 3, contoh 4,

B. 2 Rumus Jumlah dan Selisih

• rumus jumlah dan selisih 1
• rumus jumlah dan selisih 2
• materi terbaru : rumus jumlah dan selisih dudut sinus, cosinus, tangen, rumus sudut rangkap/ganda, sudut paruh/tengahan, rumus perkalian sinus dan cosinus, rumus jumlah dan selisih sinus dan cosinus, 
• soal persiapan menghadapi semester gasal: contoh 1, contoh 2, contoh 3, contoh 4, contoh 5, contoh 6, contoh 7.

B. 3 Persamaan Lingkaran

• materi lingkaran, kedudukan titik terhadap lingkaran, kedudukan garis terhadap lingkaran, persamaan garis singgung lingkaran, atau Persamaan Garis singgung Lingkaran (Lengkap), Persamaan Garis Singgung dengan gradien m, Persamaan Garis Singgung melalui sebuah titik di luar lingkaran, hubungan dua lingkaran
• contoh soal 1, contoh  2, contoh 3, contoh 4. Contoh 5 (Hubungan Dua Lingkaran), Contoh 6, Contoh 7, Contoh 8

B. 4 Polinom

• materi pendahuluan, operasi polinom, metode Horner-Kino (materi lanjutan operasi polinom) teorema sisa dan teorema faktor, persamaan polinom
• contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5, contoh 6, contoh 7, contoh 8, contoh 9

C. Kelas XII (Duabelas)

C. 1 Limit Fungsi Trigonometri

• limit fungsi trigonometri (materi lama-perkenalan dengan limit fungsi aljabar), lanjutan materi lama, lanjutan rumus dasar limit trigonometri, 
• materi pendahuluan limit, menentukan nilai limit fungsi trigonometri, pembuktian teorema apit (identitas limit fungsi trigonometri), limit fungsi trigonometri di tak hingga,
• contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5,
• materi limit diketakhinggan.
• contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5,

C.2 Turunan Fungsi Trigonometri

• materi turunan fungsi trigonometri
• lanjutan materi 
• contoh soal  (bagaian 1)
• lanjutan materi 2 
• lanjutan materi 3
• lanjutan materi 4 (sifat-sifat turunan)
• lanjutan materi 5
• contoh soal 1 (bagain 2) , contoh 2 , contoh 3 contoh 4 , contoh 5
• lanjutan materi 6 (persamaan garis singgung kurva)-turunan pertama
• lanjutan materi 7 (fungsi naik dan fungsi turun)-turunan pertama
• lanjutan materi 8 (nilai stasioner)-turunan pertama
• lanjutan materi 9 (aplikasi nilai stasioner)-turunan pertama
• contoh 6 , contoh 7 , contoh 8 , contoh 9
• turunan kedua fungsi trigonometri (Fungsi naik dan fungsi turun, titik belok serta selang kecekungan)
• contoh 10 , contoh 11 , contoh 12 , contoh 13

C.3 Distribusi peluang binomial

• materi distribusi binomial, distribusi peluang diskrit, 
• contoh soal 1, contoh 2, 
• distribusi peluang kontinu, Binomial Newton,
• contoh soal 1, contoh 2, contoh 3,

C.4 Distribusi normal

• materi distribusi normal, lanjutan 1 materi, lanjutan 2 materi
• materi distribusi student, penarikan kesimpulan (pengujian hipotesis)
• contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5, contoh 6