Answer
$\displaystyle \frac{5\sqrt{3}+\sqrt{6}}{23}=\frac{(5+\sqrt{2})\sqrt{3}}{23}$
Work Step by Step
We rationalize the denominator:
$\displaystyle \frac{\sqrt{3}}{5-\sqrt{2}}=\frac{\sqrt{3}(5+\sqrt{2})}{(5-\sqrt{2})(5+\sqrt{2})}=\frac{5\sqrt{3}+\sqrt{3}\sqrt{2}}{5^2-2}=\frac{5\sqrt{3}+\sqrt{6}}{23}$
