Question

$\sqrt{72}$
b)$\sqrt{48}$
c)$\sqrt{75}$
d)$\sqrt{90$
e)$\frac{ \sqrt{}54 }{3}$
f)$\frac{ \sqrt{1} }{125}$
g)$\frac{1}{8} \sqrt{96}$
h)$\frac{1}{7} \sqrt{147}$ scoaterea factorului de sub radical

Answer 1

          
$a)\,\, \sqrt{72} = \sqrt{2*36} = 6 \sqrt{2} \\ \\ b)\,\,\sqrt{48} = \sqrt{16 * 3}= 4\sqrt{3} \\ \\ c)\,\, \sqrt{75}= \sqrt{25*3}=5 \sqrt{3} \\ \\ d)\,\, \sqrt{90}= \sqrt{9*10}=3 \sqrt{10}$

  
$e)\,\, \frac{ \sqrt{54} }{3} = \frac{ \sqrt{9*6} }{3}= \frac{ 3\sqrt{6} }{3}= \sqrt{6} \\ \\ f)\,\, \sqrt{ \frac{1}{125}}= \frac{1}{ \sqrt{125}}= \frac{1}{ \sqrt{25*5}}=\frac{1}{ 5\sqrt{5}}=\frac{ \sqrt{5} }{ 5*5}= \frac{ \sqrt{5}}{25} \\ \\ g)\,\, \frac{1}{8} \sqrt{96}= \frac{ \sqrt{96} }{8}= \frac{ \sqrt{16*6} }{8}= \frac{ 4\sqrt{6} }{8}= \frac{ \sqrt{6} }{2} \\ \\ h)\,\, \frac{1}{7} \sqrt{147}=\frac{\sqrt{147}}{7} = \frac{\sqrt{49*3}}{7} =\frac{7\sqrt{3}}{7}= \sqrt{3}$