Removing the root squares from this expression? 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?
 A: 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}
A: 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.
A: $\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}=\frac {\sqrt{2}+ \sqrt{3}- \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}=\frac{\sqrt{2}+ \sqrt{3} - \sqrt5}{2\sqrt{6}}$ I think you can do the rest.
A: Given expression $$\frac{1}{\sqrt 2+\sqrt 3+\sqrt 5}=\frac{\sqrt 5+\sqrt 3-\sqrt 2}{(\sqrt 5+\sqrt 3)^2-2}\\=\frac{\sqrt 5+\sqrt 3-\sqrt 2}{6+2\sqrt{15}}=\frac{(\sqrt 5+\sqrt 3-\sqrt 2)(6-2\sqrt{15})}{-24}$$
