TENTUKANLAH HASIL PERKALIAN DAN PEMBAGIAN BENTUK AKAR BERIKUT !1. 3 / √5 2. 2 / 2√53. 8 / √5 + 2 4. .... Question from @kevinziggy93

September 2022 1 9 Report

TENTUKANLAH HASIL PERKALIAN DAN PEMBAGIAN BENTUK AKAR BERIKUT !

1. 3 / √5
2. 2 / 2√5
3. 8 / √5 + 2
4. √6 ( 5 + √12 )
5. √18 / √2
6. 2√5 / 3 + √5
7. 2√5 + 2√3 / 2√5 – 2√3

YANG BISA JAWAB AKU KASIH JAWABANTERCEDAS

PAKE CARA !

JΟY

Hasil perkalian dan pembagian bentuk akar:

$\begin{aligned}&{\sf{1.~\frac{3}{\sqrt{\sf5}}=\boldsymbol{\frac{\textsf{\textbf{3}}\sqrt{\textsf{\textbf{5}}}}{\textsf{\textbf{5}}}}}}\\&{\sf{2.~\frac{2}{2\sqrt{\sf5}}=\boldsymbol{\frac{\sqrt{\textsf{\textbf{5}}}}{\textsf{\textbf{5}}}}}}\\&{\sf{3.~\frac{8}{\sqrt{\sf5}+2}=\boldsymbol{\textsf{\textbf{8}}\sqrt{\textsf{\textbf{5}}}-\textsf{\textbf{16}}}}}\\&{\sf{4.~\sqrt{\sf6}\left(5+\sqrt{\sf12}\right)}=\boldsymbol{\textsf{\textbf{5}}\sqrt{\textsf{\textbf{6}}}+\textsf{\textbf{6}}\sqrt{\textsf{\textbf{2}}}}}\\&{\sf{5.~\dfrac{\sqrt{\sf18}}{\sqrt{\sf2}}=\textsf{\textbf{3}}}}\\&{\sf{6.~\dfrac{2\sqrt{\sf5}}{3+\sqrt{\sf5}}=\boldsymbol{\frac{\textsf{\textbf{3}}\sqrt{\textsf{\textbf{5}}}-\textsf{\textbf{5}}}{\textsf{\textbf{2}}}}}}\\&{\sf{7.~\frac{2\sqrt{\sf5}+2\sqrt{\sf3}}{2\sqrt{\sf5}-2\sqrt{\sf3}}=\boldsymbol{\textsf{\textbf{4}}+\sqrt{\textsf{\textbf{15}}}}}}\end{aligned}$

Pembahasan

–› No. 1

$\begin{aligned}{\sf{\frac{3}{\sqrt{\sf5}}}}&={\sf{\frac{3}{\sqrt{\sf5}}\times\frac{\sqrt{\sf5}}{\sqrt{\sf5}}}}\\&={\sf{\frac{3\sqrt{\sf5}}{\sqrt{\sf5}\sqrt{\sf5}}}}\\&={\sf{\frac{3\sqrt{\sf5}}{5}}}\end{aligned}$

–› No. 2

$\begin{aligned}{\sf{\frac{2}{2\sqrt{\sf5}}}}&={\sf{\frac{2}{2\sqrt{\sf5}}\times\frac{2\sqrt{\sf5}}{2\sqrt{\sf5}}}}\\&={\sf{\frac{2\left(2\sqrt{\sf5}\right)}{\left(2\sqrt{\sf5}\right)\left(2\sqrt{\sf5}\right)}}}\\&={\sf{\frac{2\left(2\sqrt{\sf5}\right)}{\left(4\right)\left(5\right)}}}\\&={\sf{\frac{\cancel4\sqrt{\sf5}}{\cancel{20}}}}\\&={\sf{\frac{\sqrt{\sf5}}{5}}}\end{aligned}$

–› No. 3

$\begin{aligned}{\sf{\frac{8}{\sqrt{\sf5}+2}}}&={\sf{\frac{8}{\sqrt{\sf5}+2}\times\frac{\sqrt{\sf5}-2}{\sqrt{\sf5}-2}}}\\&={\sf{\frac{8\left(\sqrt{\sf5}-2\right)}{\left(\sqrt{\sf5}+2\right)\left(\sqrt{\sf5}-2\right)}}}\\&={\sf{\frac{8\left(\sqrt{\sf5}-2\right)}{5-4}}}\\&={\sf{\frac{8\left(\sqrt{\sf5}-2\right)}{1}}}\\&={\sf{8\sqrt{\sf5}-16}}\end{aligned}$

–› No. 4

$\begin{aligned}{\sf{\sqrt{\sf6}\left(5+\sqrt{\sf12}\right)}}&={\sf{\sqrt{\sf6}\left(5+\sqrt{\sf4\cdot3}\right)}}\\&={\sf{\sqrt{\sf6}\left(5+\sqrt{\sf2^2\cdot3}\right)}}\\&={\sf{\sqrt{\sf6}\left(5+2\sqrt{\sf3}\right)}}\\&={\sf{5\sqrt{\sf6}+\sqrt{\sf18}}}\\&={\sf{5\sqrt{\sf6}+2\sqrt{\sf9\cdot2}}}\\&={\sf{5\sqrt{\sf6}+2\sqrt{\sf3^2\cdot2}}}\\&={\sf{5\sqrt{\sf6}+6\sqrt{\sf2}}}\end{aligned}$

–› No. 5

$\begin{aligned}\frac{\sqrt{\sf18}}{\sqrt{\sf2}}&=\sqrt{\frac{\sf18}{\sf2}}\\&=\sqrt{\sf9}\\&=\sqrt{\sf3^2}\\&={\sf{3}}\end{aligned}$

–› No. 6

$\begin{aligned}{\sf{\frac{2\sqrt{\sf5}}{3+\sqrt{\sf5}}}}&={\sf{\frac{2\sqrt{\sf5}}{3+\sqrt{\sf5}}\times\frac{3-\sqrt{\sf5}}{3-\sqrt{\sf5}}}}\\&={\sf{\frac{2\sqrt{\sf5}\left(3-\sqrt{\sf5}\right)}{\left(3+\sqrt{\sf5}\right)\left(3-\sqrt{\sf5}\right)}}}\\&={\sf{\frac{2\sqrt{\sf5}\left(3-\sqrt{\sf5}\right)}{9-5}}}\\&={\sf{\frac{\cancel2\sqrt{\sf5}\left(3-\sqrt{\sf5}\right)}{\cancel4}}}\\&={\sf{\frac{\sqrt{\sf5}\left(3-\sqrt{\sf5}\right)}{2}}}\\&={\sf{\frac{3\sqrt{\sf5}-5}{2}}}\end{aligned}$

–› No. 7

$\begin{aligned}{\sf{\frac{2\sqrt{\sf5}+2\sqrt{\sf3}}{2\sqrt{\sf5}-2\sqrt{\sf3}}}}&={\sf{\frac{2\sqrt{\sf5}+2\sqrt{\sf3}}{2\sqrt{\sf5}-2\sqrt{\sf3}}\times{\sf{\frac{2\sqrt{\sf5}+2\sqrt{\sf3}}{2\sqrt{\sf5}+2\sqrt{\sf3}}}}}}\\&={\sf{\frac{\left(2\sqrt{\sf5}+2\sqrt{\sf3}\right)\left(2\sqrt{\sf5}+2\sqrt{\sf3}\right)}{\left(2\sqrt{\sf5}-2\sqrt{\sf3}\right)\left(2\sqrt{\sf5}+2\sqrt{\sf3}\right)}}}\\&={\sf{\frac{32+8\sqrt{15}}{8}}}\\&={\sf{\frac{\cancel8\left(4+\sqrt{15}\right)}{\cancel8}}}\\&={\sf{4+\sqrt{15}}}\end{aligned}$