How to show $\frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2} - \sqrt{3}}{5}$? Show that:
$$ \dfrac{1}{2\sqrt2+\sqrt3}=\dfrac{2\sqrt2-\sqrt3}{5}$$
So I multiplied everything by $\sqrt3$
Then I got 
$$\frac{\sqrt{3}}{2\sqrt{2}+3}$$
Then multiply it by $\sqrt2$ to obtain
$$\frac{\sqrt{2}\sqrt{3}}{2 \cdot 3+3}$$
Which is $$\frac{\sqrt{2}\sqrt{3}}{9}$$ which isn't equal to $$\frac{2\sqrt{2}-\sqrt{3}}{5}$$
What did I do wrong?

\begin{align}
\frac {1}{2\sqrt 2 + \sqrt 3}= \frac {2\sqrt 2-\sqrt 3}{5}&\iff \\\frac {1}{2\sqrt 2 + \sqrt 3}\cdot  {2\sqrt 2 + \sqrt 3} = \frac {2\sqrt 2-\sqrt 3}{5}\cdot  {(2\sqrt 2 + \sqrt 3)} &\iff\\
1=\frac {(2\sqrt2)^2-(\sqrt 3)^2}{5}&\iff\\
1=\frac {4\cdot 2-3}{5}&\iff 1=\frac 5 5.
\end{align}
