Question

Racionalização de denominadores:
$\frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } = \frac{ \sqrt{2}- \sqrt{3} }{ \sqrt{2} - \sqrt{3} } .$$\frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} }$

Answer 1

$\dfrac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\\ \\ \\ =\dfrac{\sqrt{6}\cdot (\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})\cdot (\sqrt{2}-\sqrt{3})}\\ \\ \\ =\dfrac{\sqrt{6}\cdot \sqrt{2}-\sqrt{6}\cdot \sqrt{3}}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}\\ \\ \\ =\dfrac{\sqrt{6\cdot 2}-\sqrt{6\cdot 3}}{2-3}\\ \\ \\ =\dfrac{\sqrt{12}-\sqrt{18}}{-1}\\ \\ \\ =\sqrt{18}-\sqrt{12}\\ \\ =\sqrt{9\cdot 2}-\sqrt{4\cdot 3}\\ \\ =\sqrt{9}\cdot\sqrt{2}-\sqrt{4}\cdot \sqrt{3}\\ \\ =3\sqrt{2}-2\sqrt{3}$