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

$$\begin{array}{rl}{\displaystyle \frac{\sqrt{5}+\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}}& ={\displaystyle \frac{\sqrt{5}+\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}}\cdot {\displaystyle \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}}\\ & ={\displaystyle \frac{\sqrt{10}+\sqrt{15}+5+\sqrt{6}+3+\sqrt{15}-2-\sqrt{6}-\sqrt{10}}{{(\sqrt{2}+\sqrt{3})}^2-5}}\\ & ={\displaystyle \frac{2\sqrt{15}+6}{2+2\sqrt{6}+3-5}}\\ & ={\displaystyle \frac{2\sqrt{15}+6}{2\sqrt{6}}}\\ & ={\displaystyle \frac{\sqrt{15}+3}{\sqrt{6}}}\cdot {\displaystyle \frac{\sqrt{6}}{\sqrt{6}}}\\ & ={\displaystyle \frac{\sqrt{90}+3\sqrt{6}}{6}}\\ & ={\displaystyle \frac{3\sqrt{10}+3\sqrt{6}}{6}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{\sqrt{10}+\sqrt{6}}{2}}}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \frac{2\sqrt{3}+4\sqrt{27}-4\sqrt{3}}{\sqrt{2}}}& ={\displaystyle \frac{2\sqrt{3}+4\cdot 3\sqrt{3}-4\sqrt{3}}{\sqrt{2}}}\\ & ={\displaystyle \frac{2\sqrt{3}+12\sqrt{3}-4\sqrt{3}}{\sqrt{2}}}\\ & ={\displaystyle \frac{10\sqrt{3}}{\sqrt{2}}}\cdot {\displaystyle \frac{\sqrt{2}}{\sqrt{2}}}\\ & ={\displaystyle \frac{10\sqrt{6}}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 5\sqrt{6}}}}}\end{array}$$

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
   

$$\begin{array}{rl}-4\sqrt{200}+2\sqrt{242}+5\sqrt{50}-10\sqrt{2}& =-4\cdot 10\sqrt{2}+2\cdot 11\sqrt{2}+5\cdot 5\sqrt{2}-10\sqrt{2}\\ & =-40\sqrt{2}+22\sqrt{2}+25\sqrt{2}-10\sqrt{2}\\ & =-3\sqrt{2}\end{array}$$

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

$$\begin{array}{rl}r& ={\displaystyle \frac{20\sqrt{2}-25}{(10+20\sqrt{2})(2-\sqrt{2})}}\\ & ={\displaystyle \frac{5(4\sqrt{2}-5)}{10(1+2\sqrt{2})(2-\sqrt{2})}}\\ & ={\displaystyle \frac{4\sqrt{2}-5}{2(1+2\sqrt{2})(2-\sqrt{2})}}\\ & ={\displaystyle \frac{4\sqrt{2}-5}{2(2-\sqrt{2}+4\sqrt{2}-4)}}\\ & ={\displaystyle \frac{24+8\sqrt{2}-15\sqrt{2}-10}{2(18-4)}}\\ & ={\displaystyle \frac{14-7\sqrt{2}}{2(14)}}\\ & ={\displaystyle \frac{2-\sqrt{2}}{2(2)}}\\ & ={\displaystyle \frac{2-\sqrt{2}}{4}}\end{array}$$

$$\begin{array}{rl}(4r-2{)}^2& ={(4\left({\displaystyle \frac{2-\sqrt{2}}{4}}\right)-2)}^2\\ & ={(2-\sqrt{2}-2)}^2\\ & ={(-\sqrt{2})}^2\\ & =2\end{array}$$

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

No.

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

GANESHA OPERATION

$$\begin{array}{rl}\sqrt{237}& =\sqrt{2,37\cdot 100}\\ & =\sqrt{2,37}\cdot 10\\ & =3p\cdot 10\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 30p}}}}\end{array}$$

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
   

$$\begin{array}{rl}{\displaystyle \frac{5-5\sqrt{2}}{\sqrt{5}-\sqrt{10}}}& ={\displaystyle \frac{5(1-\sqrt{2})}{\sqrt{5}(1-\sqrt{2})}}\\ & ={\displaystyle \frac{5}{\sqrt{5}}}\cdot {\displaystyle \frac{\sqrt{5}}{\sqrt{5}}}\\ & ={\displaystyle \frac{5\sqrt{5}}{5}}\\ b& =\sqrt{5}\end{array}$$

$$\begin{array}{rl}{}^b\log 125& ={}^{\sqrt{5}}\log 125\\ & ={}^{5^{\frac{1}{2}}}\log 5^3\\ & =2\cdot 3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 6}}}}\end{array}$$

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

$$\begin{array}{rl}\sqrt{3+2\sqrt{2}}-\sqrt{2}& =\sqrt{2+1+2\sqrt{2\cdot 1}}-\sqrt{2}\\ & =\cancel{\sqrt{2}}+\sqrt{1}-\cancel{\sqrt{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$$

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

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

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

No.

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

Telegram

$$\begin{array}{rl}(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})& =7-5\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$$

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

$$\begin{array}{rl}\sqrt{108}+\sqrt{15}\times \sqrt{5}-2\sqrt{48}& =\sqrt{36\cdot 3}+\sqrt{3\cdot 5}\times \sqrt{5}-2\sqrt{16\cdot 3}\\ & =6\sqrt{3}+\sqrt{3}\times \sqrt{5}\times \sqrt{5}-2\times 4\sqrt{3}\\ & =6\sqrt{3}+5\sqrt{3}-8\sqrt{3}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3\sqrt{3}}}}}\end{array}$$

Jadi, JAWAB: