TITLE: Removing the root squares from this expression?
QUESTION [5 upvotes]: I would like to understand how to remove the root squares from this expression:
$$x = \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}$$
How to do it?

REPLY [7 votes]: Its all about rationalization,
\begin{align}
\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}} &= \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}
\cdot \frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} - \sqrt{5}}
\\[10pt] &=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}
\\[10pt] &=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{(2+3+2\sqrt{6})-(5)}
\\[10pt] &=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{2\sqrt{6}}\cdot\frac{\sqrt6}{\sqrt6}
\\[10pt]&=\frac{\sqrt{12}+ \sqrt{18} - \sqrt{30}}{12}
\\[10pt]&=\frac{2\sqrt{3} + 3\sqrt{2} - \sqrt{30}}{12}
\end{align}

REPLY [3 votes]: Here's a push
You can try rationalising it by multiplying the expression by its conjugate.
$$\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}\times\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} - \sqrt{5}}=\frac {\sqrt{2}+ \sqrt{3}- \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}$$
and you can continue it.