<s>[INST] 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?
 [/INST] The conjugate of $\sqrt{a} + \sqrt{b}$ is $\sqrt{a} - \sqrt{b}$.  To rationalize an expression of the form $$\frac{1}{\sqrt{a} + \sqrt{b}}$$ we multiply the numerator and denominator by $\sqrt{a} - \sqrt{b}$.  Since the conjugate of $2\sqrt{2} + \sqrt{3}$ is $2\sqrt{2} - \sqrt{3}$, we obtain
\begin{align*}
\frac{1}{2\sqrt{2} + \sqrt{3}} & = \frac{1}{2\sqrt{2} + \sqrt{3}} \cdot \frac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} - \sqrt{3}} && \text{multiply by the conjugate}\\
                               & = \frac{2\sqrt{2} - \sqrt{3}}{(2\sqrt{2})^2 - (\sqrt{3})^2} && \text{since $(a + b)(a - b) = a^2 - b^2$}\\
                               & = \frac{2\sqrt{2} - \sqrt{3}}{8 - 3}\\
                               & = \frac{2\sqrt{2} - \sqrt{3}}{5}
\end{align*}
As for your mistakes, you did not apply the distributive law when you multiplied the numerator and denominator by $\sqrt{3}$.  You should have obtained 
$$\frac{1}{2\sqrt{2} + \sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{2\sqrt{2}\sqrt{3} + \sqrt{3}\sqrt{3}} = \frac{\sqrt{3}}{2\sqrt{6} + 3}$$
You also failed to distribute the factor of $\sqrt{2}$ in the denominator in the next step.  Once you obtained $$\frac{\sqrt{3}}{2\sqrt{6} + 3}$$ you could have multiplied the numerator and denominator by the conjugate of the denominator, which is $-2\sqrt{6} + 3$, then simplified.  
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