# Simplifying a square root fraction

Simplify the following

$$\frac{\sqrt{3}}{\sqrt{2}(\sqrt{6} - \sqrt{3})}$$

Apparently the answer is $\frac{1}{2} (2 + \sqrt{2})$ but can't for the life of me see how to get it. Any help is massively appreciated.

Thanks

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$\dfrac{\sqrt 3}{\sqrt 2 (\sqrt 6 - \sqrt 3)} = \dfrac{\sqrt 3}{\sqrt 2 \sqrt 3(\sqrt2 - 1)} = \dfrac{1}{\sqrt 2 (\sqrt 2 - 1)} = \dfrac{1}{(2 - \sqrt 2)} = \dfrac{(2 + \sqrt 2)}{(2 - \sqrt 2)(2 + \sqrt 2)} = \dfrac{2 + \sqrt 2}{2}$
The reason we write it this way is that it is bad practice to have roots at the denominator. To get rid of $a+b \sqrt c$ at the denominator, simply multiply the numerator and the denominator by $a-b \sqrt c$.
$\frac{\sqrt{3}}{\sqrt{2}(\sqrt{6}-\sqrt{3})}=\frac{\sqrt{1}}{\sqrt{2}(\sqrt{2}-\sqrt{1})}=\frac{\sqrt{2}+\sqrt{1}}{\sqrt{2}}=\frac12(2+\sqrt{2}).$