$\frac{2 căn 3 -3 căn 2}{căn 6}$- $\frac{2- căn 2}{1- căn 2}$ +$\frac{3}{căn 3}$ giúp em bài này với ạ

$\dfrac{2\sqrt{3}-3\sqrt{2}}{\sqrt{6}}-\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{3}{\sqrt{3}}$

$=\dfrac{2-\sqrt{6}}{\sqrt{2}}-\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{3}{\sqrt{3}}$

$=\dfrac{\left(2-\sqrt{6}\right)\sqrt{3}\left(1-\sqrt{2}\right)}{\sqrt{6}\left(1-\sqrt{2}\right)}-\dfrac{\left(2-\sqrt{2}\right)\sqrt{6}}{\sqrt{6}\left(1-\sqrt{2}\right)}+\dfrac{3\left(\sqrt{2}-2\right)}{\sqrt{6}\left(1-\sqrt{2}\right)}$

$=\dfrac{\left(2-\sqrt{6}\right)\sqrt{3}\left(1-\sqrt{2}\right)-\left(2-\sqrt{2}\right)\sqrt{6}+3\left(\sqrt{2}-2\right)}{\sqrt{6}\left(1-\sqrt{2}\right)}$

$=\dfrac{4\sqrt{3}-4\sqrt{6}}{\sqrt{6}\left(1-\sqrt{2}\right)}$

$=-\dfrac{-2\sqrt{6}}{\sqrt{3}}$

$=2\sqrt{2}$