Show that :

$\dfrac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \dfrac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ = 11

Given,

Equation : $\dfrac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \dfrac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ = 11

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \dfrac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \dfrac{\sqrt{3}\sqrt{2}(\sqrt{3} - \sqrt{2})}{\sqrt{3}\sqrt{2}(\sqrt{3} + \sqrt{2})} + \dfrac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} + \dfrac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \dfrac{(\sqrt{3} - \sqrt{2})^2 + 2\sqrt{3}(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{(\sqrt{3})^2 + (\sqrt{2})^2 - 2\times \sqrt{3} \times \sqrt{2} + 6 + 2\sqrt{6}}{3 - 2} \\[1em] \Rightarrow \dfrac{3 + 2 - 2\sqrt{6} + 6 + 2\sqrt{6}}{1} \\[1em] \Rightarrow 11.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \dfrac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ = 11.