Question

Como resolver

$\frac{3}{ \sqrt{3} +\sqrt{2} }$

$\frac{6}{3- \sqrt{2} }$

Answer 1

$\frac{3}{\sqrt{3}+\sqrt{2}} . \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{3.(\sqrt{3}-\sqrt{2})}{3-2} =3.(\sqrt{3}-\sqrt{2}) \\ \frac{6}{3-\sqrt{2}} .\frac{3+\sqrt{2}}{3+\sqrt{2}}= \frac{6.(3+\sqrt{2})}{9-2} =\frac{6.(3+\sqrt{2})}{7}$