Exercise Zone : Bentuk Akar

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Tipe:

Standar SBMPTN HOTS

No.

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

Grup Matematika Zone Indonesia

\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

No.

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

Grup Matematika Zone Indonesia

\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.

No.

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

Grup Matematika Zone Indonesia

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

No.

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
   

Grup Matematika Zone Indonesia

\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

No.

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

No.

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}

\begin{aligned} ^b\negmedspace\log125&={^{\sqrt5}\negmedspace\log}125\\ &={^{5^{\frac12}}\negmedspace\log}5^3\\
&=2\cdot3\\ &=\boxed{\boxed{6}} \end{aligned}

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

No.

{\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

No.

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)

Zenius.net

\begin{aligned} \dfrac{\left(\sqrt3+\sqrt7\right)\left(\sqrt3-\sqrt7\right)}{2\sqrt5-4\sqrt2}&=\dfrac{3-7}{2\sqrt5-4\sqrt2}{\color{red}{\times\dfrac{2\sqrt5+4\sqrt2}{2\sqrt5+4\sqrt2}}}\\[4pt] &=\dfrac{-4\left(2\sqrt5+4\sqrt2\right)}{20-32}\\[4pt] &=\dfrac{-4\left(2\sqrt5+4\sqrt2\right)}{-12}\\[4pt] &=\dfrac{2\sqrt5+4\sqrt2}3\\[4pt] &=\dfrac{2\left(\sqrt5+2\sqrt2\right)}3\\ &=\boxed{\boxed{\dfrac23\left(\sqrt5+2\sqrt2\right)}} \end{aligned}

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

No.

Tentukan nilai dari \left(\sqrt7+\sqrt5\right)\left(\sqrt7-\sqrt5\right)

Telegram

\begin{aligned} \left(\sqrt7+\sqrt5\right)\left(\sqrt7-\sqrt5\right)&=7-5\\ &=\boxed{\boxed{2}} \end{aligned}

Jadi, \left(\sqrt7+\sqrt5\right)\left(\sqrt7-\sqrt5\right)=2.

No.

Hasil dari {\sqrt{108}+\sqrt{15}\times\sqrt5-2\sqrt{48}} adalah

1. -3\sqrt{15}
2. \sqrt3

3. 3\sqrt3
4. 3\sqrt{15}

SKMP Januari 2021

\begin{aligned} \sqrt{108}+\sqrt{15}\times\sqrt5-2\sqrt{48}&=\sqrt{36\cdot3}+\sqrt{3\cdot5}\times\sqrt5-2\sqrt{16\cdot3}\\ &=6\sqrt3+\sqrt3\times\sqrt5\times\sqrt5-2\times4\sqrt3\\ &=6\sqrt3+5\sqrt3-8\sqrt3\\ &=\boxed{\boxed{3\sqrt3}} \end{aligned}

Jadi, JAWAB: