Rationalizing mixed denominators? How would I rationalize the following Fraction?
$$ \frac {2}{5-\sqrt2+\sqrt3}$$
I have considered the idea of multiplying by the same radicals, but the 5 prevents that.
 A: Multiply top and bottom by all the "relatives" $5+\sqrt{2}-\sqrt{3}$, $5+\sqrt{2}+\sqrt{3}$ and $5-\sqrt{2}-\sqrt{3}$. 
The new denominator is invariant under replacement of $\sqrt{2}$ by $-\sqrt{2}$, also under replacement of $\sqrt{3}$ by $-\sqrt{3}$, so it must be rational. 
A: Hint $\ $ Rationalize the denominator using the product of the terms below
$\quad \begin{eqnarray}(5\!+\!\sqrt3-\sqrt2)(5\!+\!\sqrt3+\sqrt2) &\,=\,& (5\!+\!\sqrt3)^2-2 &\,=\,&26+10\sqrt3\\  
(5\!-\!\sqrt3-\sqrt2)(5\!-\!\sqrt3+\sqrt2) &=& (5\!-\!\sqrt3)^2-2  &=&26-10\sqrt3\end{eqnarray}\Bigg\rbrace$ multiplied $ \,=\, \ldots$
A: $$\begin{align}\frac{2}{5-\sqrt2+\sqrt3}\left(\frac{5+\sqrt2-\sqrt3}{5+\sqrt2-\sqrt3}\right)&=\frac{10+2\sqrt2-2\sqrt3}{20+2\sqrt6}\\
&=\frac{10+2\sqrt2-2\sqrt3}{20+2\sqrt6}\left(\frac{20-2\sqrt6}{20-2\sqrt6}\right)\\
&=\frac{200-20\sqrt6+40\sqrt2-4\sqrt{12}-40\sqrt3+4\sqrt{18}}{400-24}\\
&=\frac{200-20\sqrt6+52\sqrt2-48\sqrt3}{376}\\
&=\frac{50-5\sqrt6+13\sqrt2-12\sqrt3}{94}\\
&\approx.37609
\end{align}$$
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