Question

1÷√(8-√32)
There is the root of whole bracket

Answer 1

1 ÷ $\sqrt{(8-\sqrt{32})}$$=\dfrac{1}{2\sqrt{2-\sqrt{2}}}$

Step-by-step explanation:

We have,

1 ÷ $\sqrt{(8-\sqrt{32})}$

To find, the value of 1 ÷ $\sqrt{(8-\sqrt{32})}$ =?

∴ 1 ÷ $\sqrt{(8-\sqrt{32})}$

$=1 \times \dfrac{1}{\sqrt{8-\sqrt{32}}}$

$=1 \times \dfrac{1}{\sqrt{8-\sqrt{16\times 2}}}$

$=\dfrac{1}{\sqrt{4\times 2-4\sqrt{2}}}$

$=\dfrac{1}{2\sqrt{2-\sqrt{2}}}$

Hence, 1 ÷ $\sqrt{(8-\sqrt{32})}$$=\dfrac{1}{2\sqrt{2-\sqrt{2}}}$