Question

simplify root 6/root 2+root+3 root 2/root6+root3-4root3/root6+root2​

Answer 1

$\bold{Expression : \frac{6}{2\sqrt3-\sqrt6}+\frac{\sqrt6}{\sqrt3+\sqrt2}-\frac{4\sqrt3}{\sqrt6-\sqrt2} }$

Rationalize each term,

$\bold{=\frac{6}{2\sqrt3-\sqrt6}\times \frac{2\sqrt3+\sqrt6}{2\sqrt3+\sqrt6}+\frac{\sqrt6}{\sqrt3+\sqrt2}\times \frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}-\frac{4\sqrt3}{\sqrt6-\sqrt2}\times \frac{\sqrt6+\sqrt2}{\sqrt6+\sqrt2}}$

$\bold{=\frac{6(2\sqrt3+\sqrt6)}{12-6}+\frac{\sqrt6(\sqrt3-\sqrt2)}{3-2}-\frac{4\sqrt3(\sqrt6+\sqrt2)}{6-2}}$

$\bold{=\frac{6(2\sqrt3+\sqrt6)}{6}+\frac{\sqrt6(\sqrt3-\sqrt2)}{1}-\frac{4\sqrt3(\sqrt6+\sqrt2)}{4}}$

$\bold{=2\sqrt3+\sqrt6+3\sqrt 2-2\sqrt3-3\sqrt2-\sqrt6} = 0$

$\bold{Therefore, \frac{6}{2\sqrt3-\sqrt6}+\frac{\sqrt6}{\sqrt3+\sqrt2}-\frac{4\sqrt3}{\sqrt6-\sqrt2}=0}$