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?
 A: $$(2\sqrt2-\sqrt3)(2\sqrt2+\sqrt3)=(2\sqrt2)^2-(\sqrt3)^2=?$$
A: As one of the comments stated, your first error lies in the first step where you multiply "everything" by $\sqrt{3}$. While you claim to do this, you did not multiply both terms in the denominator by $\sqrt{3}$. This was an error that let you to an incorrect expression.
That having been said, your approach is not correct. To show this equality, you have to simplify the expression on the left hand side. In this case, because you have radical expressions in the denominator, that means multiplying the top and bottom by the "right value" which will eliminate the radicals. In the case of a single term in the denominator like $\dfrac{1}{\sqrt{2}}$, you would simply multiply the top and bottom by $\sqrt{2}$ which would yield $$\dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$$
The case you have, however, contains a binomial in the denominator (meaning two terms). In this case, you should multiply by the conjugate of the denominator. So for your example, the conjugate of $2\sqrt{2} + \sqrt{3}$ would be $2\sqrt{2} - \sqrt{3}$. Doing so yields the following:
\begin{align}
   \frac{1}{2\sqrt{2} + \sqrt{3}} \cdot \frac{2\sqrt{2}-\sqrt{3}}{2\sqrt{2} - \sqrt{3}} &= \frac{2\sqrt{2}-\sqrt{3}}{(2\sqrt{2})^2 - (\sqrt{3})^2} \\
&= \frac{2\sqrt{2}-\sqrt{3}}{8 - 3} \\
&= \frac{2\sqrt{2}-\sqrt{3}}{5}
\end{align}
Thus we've shown that $$ \frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2}-\sqrt{3}}{5}$$
A: 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.  
A: \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}
