Answer
$\displaystyle \frac{3\sqrt{6}+4}{2}$
Work Step by Step
We lose the square roots in the denominator by applying the difference of squares formula:
$(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$
$\displaystyle \frac{5\sqrt{3}-3\sqrt{2}}{3\sqrt{2}-2\sqrt{3}}\color{red}{ \cdot\frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} }\qquad$ (rationalize)
$=\displaystyle \frac{(5\sqrt{3}-3\sqrt{2})(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2})^{2}-(2\sqrt{3})^{2}}$
... use FOIL for the numerator
$=\displaystyle \frac{15\sqrt{6}+10\cdot 3-9\cdot 2-6\sqrt{6}}{3^{2}\cdot 2-2^{2}\cdot 3}$
$=\displaystyle \frac{9\sqrt{6}+30-18}{18-12}$
$=\displaystyle \frac{9\sqrt{6}+12}{6}$
$=\displaystyle \frac{3(3\sqrt{6}+4)}{6}$
$=\displaystyle \frac{3\sqrt{6}+4}{2}$
