Question

Simplify 6/2root3-root6+root6/root3+root2-4root3/root6-root2

Answer 1

Answer:

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

Step-by-step explanation:

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

To find : Simplify the expression ?

Solution :

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

Rationalize each term,

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

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

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

$=2\sqrt3+\sqrt6+3\sqrt 2-2\sqrt3-3\sqrt2-\sqrt6$

$=0$

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

Answer 2

Answer:

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