Question

(1supra radical din8+ 1supra radical din 32- 1 supra radical din 2 ) : radical din 2 supra 2

Answer 1

$\displaystyle\left(\frac1{\sqrt8}+\frac1{\sqrt{32}}-\frac1{\sqrt2}\right):\frac{\sqrt2}2=\left(\frac{1}{2\sqrt2}+\frac{1}{2^2\sqrt2}-\frac{1}{\sqrt2}\right)\cdot\frac2{\sqrt2}=\frac{2+1-2^2}{2^2\sqrt2}\cdot\frac{2}{\sqrt2}=\frac{-1}{4}$