Exercise Zone : Bentuk Akar

Berikut ini adalah kumpulan soal mengenai bentuk akar tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Pecahan $\dfrac{\sqrt5+\sqrt3-\sqrt2}{\sqrt2+\sqrt3-\sqrt5}$ setara dengan....

1. $\dfrac{\sqrt{60}+\sqrt6}2$
2. $\dfrac{\sqrt{10}+\sqrt6}2$
3. $\dfrac{\sqrt{15}+\sqrt3}2$

4. $\dfrac{\sqrt{15}-\sqrt3}2$
5. $\dfrac{\sqrt{15}+\sqrt6}2$

$\begin{aligned}\dfrac{\sqrt5+\sqrt3-\sqrt2}{\sqrt2+\sqrt3-\sqrt5}&=\dfrac{\sqrt5+\sqrt3-\sqrt2}{\sqrt2+\sqrt3-\sqrt5}\cdot\dfrac{\sqrt2+\sqrt3+\sqrt5}{\sqrt2+\sqrt3+\sqrt5}\\[8pt]&=\dfrac{\sqrt{10}+\sqrt{15}+5+\sqrt6+3+\sqrt{15}-2-\sqrt6-\sqrt{10}}{\left(\sqrt2+\sqrt3\right)^2-5}\\[8pt]&=\dfrac{2\sqrt{15}+6}{2+2\sqrt6+3-5}\\[8pt]&=\dfrac{2\sqrt{15}+6}{2\sqrt6}\\[8pt]&=\dfrac{\sqrt{15}+3}{\sqrt6}\cdot\dfrac{\sqrt6}{\sqrt6}\\[8pt]&=\dfrac{\sqrt{90}+3\sqrt6}6\\[8pt]&=\dfrac{3\sqrt{10}+3\sqrt6}6\\&=\boxed{\boxed{\dfrac{\sqrt{10}+\sqrt6}2}}\end{aligned}$

Jadi, $\dfrac{\sqrt5+\sqrt3-\sqrt2}{\sqrt2+\sqrt3-\sqrt5}$ setara dengan $\dfrac{\sqrt{10}+\sqrt6}2$.
JAWAB: B

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No. 2

Bentuk sederhana dari ${\dfrac{2\sqrt3+4\sqrt{27}-4\sqrt3}{\sqrt2}}$ adalah....

$\begin{aligned}\dfrac{2\sqrt3+4\sqrt{27}-4\sqrt3}{\sqrt2}&=\dfrac{2\sqrt3+4\cdot3\sqrt3-4\sqrt3}{\sqrt2}\\[4pt]&=\dfrac{2\sqrt3+12\sqrt3-4\sqrt3}{\sqrt2}\\[4pt]&=\dfrac{10\sqrt3}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}\\[4pt]&=\dfrac{10\sqrt6}2\\[4pt]&=\boxed{\boxed{5\sqrt6}}\end{aligned}$

Jadi, $\dfrac{2\sqrt3+4\sqrt{27}-4\sqrt3}{\sqrt2}=5\sqrt6$.

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No. 3

Bentuk sederhana dari $-4\sqrt{200}+2\sqrt{242}+5\sqrt{50}-10\sqrt2$ adalah ....

1. $-3\sqrt2$
2. $-2\sqrt2$
3. $\sqrt2$

4. $2\sqrt2$
5. $3\sqrt2$

$\begin{aligned}-4\sqrt{200}+2\sqrt{242}+5\sqrt{50}-10\sqrt2&=-4\cdot10\sqrt2+2\cdot11\sqrt2+5\cdot5\sqrt2-10\sqrt2\\&=-40\sqrt2+22\sqrt2+25\sqrt2-10\sqrt2\\&=-3\sqrt2\end{aligned}$

Jadi, $-4\sqrt{200}+2\sqrt{242}+5\sqrt{50}-10\sqrt2=-3\sqrt2$.
JAWAB: A

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No. 4

Jika $r=\dfrac{20\sqrt2-25}{\left(10+20\sqrt2\right)\left(2-\sqrt2\right)}$, maka $(4r-2)^2=$ ....

1. $5$
2. $4$
   
3. $3$
   

4. $2$
5. $1$
   

$\begin{aligned}r&=\dfrac{20\sqrt2-25}{\left(10+20\sqrt2\right)\left(2-\sqrt2\right)}\\&=\dfrac{5\left(4\sqrt2-5\right)}{10\left(1+2\sqrt2\right)\left(2-\sqrt2\right)}\\&=\dfrac{4\sqrt2-5}{2\left(1+2\sqrt2\right)\left(2-\sqrt2\right)}\\&=\dfrac{4\sqrt2-5}{2\left(2-\sqrt2+4\sqrt2-4\right)}\\&=\dfrac{24+8\sqrt2-15\sqrt2-10}{2(18-4)}\\&=\dfrac{14-7\sqrt2}{2(14)}\\&=\dfrac{2-\sqrt2}{2(2)}\\&=\dfrac{2-\sqrt2}4\end{aligned}$

$\begin{aligned}(4r-2)^2&=\left(4\left(\dfrac{2-\sqrt2}4\right)-2\right)^2\\&=\left(2-\sqrt2-2\right)^2\\&=\left(-\sqrt2\right)^2\\&=2\end{aligned}$

Jadi, $(4r-2)^2=2$.
JAWAB: D

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No. 5

Jika $3p=\sqrt{2,37}$ maka nilai $\sqrt{237}$ adalah....

GANESHA OPERATION

$\begin{aligned}\sqrt{237}&=\sqrt{2,37\cdot100}\\&=\sqrt{2,37}\cdot10\\&=3p\cdot10\\&=\boxed{\boxed{30p}}\end{aligned}$

Jadi, $\sqrt{237}=30p$.
JAWAB: D

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No. 6

Jika ${\dfrac{5-5\sqrt2}{\sqrt5-\sqrt{10}}=b}$, maka $^b\negmedspace\log125=$

1. $2$
2. $3$
3. $4$

4. $5$
5. $6$

$\begin{aligned}\dfrac{5-5\sqrt2}{\sqrt5-\sqrt{10}}&=\dfrac{5\left(1-\sqrt2\right)}{\sqrt5\left(1-\sqrt2\right)}\\[8pt]&=\dfrac5{\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}\\[8pt]&=\dfrac{5\sqrt5}5\\[8pt]b&=\sqrt5\end{aligned}$

Jadi, $^b\negmedspace\log125=6$.
JAWAB: E

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No. 7

${\sqrt{3+2\sqrt2}-\sqrt2=}$ ....

1. $4\sqrt2$
2. ${3+\sqrt2}$
3. $\sqrt2$

4. $1$
5. $0$

Kemampuan Dasar SIMAK UI 2009

$\begin{aligned}\sqrt{3+2\sqrt2}-\sqrt2&=\sqrt{2+1+2\sqrt{2\cdot1}}-\sqrt2\\&=\cancel{\sqrt2}+\sqrt1-\cancel{\sqrt2}\\&=\boxed{\boxed{1}}\end{aligned}$

Jadi, $\sqrt{3+2\sqrt2}-\sqrt2=1$.
JAWAB: D

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No. 8

Bentuk sederhana dari $\dfrac{\left(\sqrt3+\sqrt7\right)\left(\sqrt3-\sqrt7\right)}{2\sqrt5-4\sqrt2}$ adalah

1. $\dfrac23\left(\sqrt5+2\sqrt2\right)$
2. $\dfrac23\left(2\sqrt2-\sqrt5\right)$
3. $-\dfrac23\left(2\sqrt5+4\sqrt2\right)$

4. $-\dfrac49\left(2\sqrt5+4\sqrt2\right)$
5. $-\dfrac49\left(2\sqrt5-\sqrt2\right)$

SUMBER

$\begin{array}{rl}{\displaystyle \frac{(\sqrt{3}+\sqrt{7})(\sqrt{3}-\sqrt{7})}{2\sqrt{5}-4\sqrt{2}}}& ={\displaystyle \frac{3-7}{2\sqrt{5}-4\sqrt{2}}}\times {\displaystyle \frac{2\sqrt{5}+4\sqrt{2}}{2\sqrt{5}+4\sqrt{2}}}\\ & ={\displaystyle \frac{-4(2\sqrt{5}+4\sqrt{2})}{20-32}}\\ & ={\displaystyle \frac{-4(2\sqrt{5}+4\sqrt{2})}{-12}}\\ & ={\displaystyle \frac{2\sqrt{5}+4\sqrt{2}}{3}}\\ & ={\displaystyle \frac{2(\sqrt{5}+2\sqrt{2})}{3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{2}{3}}(\sqrt{5}+2\sqrt{2})}}}}\end{array}$

Jadi, $\dfrac{\left(\sqrt3+\sqrt7\right)\left(\sqrt3-\sqrt7\right)}{2\sqrt5-4\sqrt2}=\dfrac23\left(\sqrt5+2\sqrt2\right)$.
JAWAB: A

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