Racionalizar o denominador de $\dfrac{2}{\sqrt{3} + 1 + \sqrt{2}}$.

$\dfrac{2}{\sqrt{3} + 1 + \sqrt{2}} = \dfrac{2[(\sqrt{3} + 1) - \sqrt{2}]}{[(\sqrt{3} + 1) + \sqrt{2}][(\sqrt{3} + 1) - \sqrt{2}]} = \dfrac{2(\sqrt{3} + 1 - \sqrt{2})]}{2 + 2\sqrt{3}} =$

 

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