Divide $\frac{\sqrt{3}-\sqrt{2}}{\frac{\sqrt{3}}{\sqrt{2}}}$ by $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ by multiplying $\frac{\sqrt{3}-\sqrt{2}}{\frac{\sqrt{3}}{\sqrt{2}}}$ by the reciprocal of $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$.
Divide $5-2\sqrt{6}$ by $\frac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{2}$ by multiplying $5-2\sqrt{6}$ by the reciprocal of $\frac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{2}$.
Rationalize the denominator of $\frac{\left(5-2\sqrt{6}\right)\times 2}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}$ by multiplying numerator and denominator by $\sqrt{6}$.
Consider $\left(6\sqrt{3}+6\sqrt{2}\right)\left(6\sqrt{3}-6\sqrt{2}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
