Question

Dúvida sobre racionalização :/

Answer 1

$E=\frac{\sqrt5-\sqrt2}{\sqrt7+\sqrt3}+\frac{\sqrt7-\sqrt3}{\sqrt5+\sqrt2}\\ \\ E=\frac{\sqrt5-\sqrt2}{\sqrt7+\sqrt3}*\frac{\sqrt7-\sqrt3}{\sqrt7-\sqrt3}+\frac{\sqrt7-\sqrt3}{\sqrt5+\sqrt2}*\frac{\sqrt5-\sqrt2}{\sqrt5-\sqrt2}\\ \\ E=\frac{\sqrt{35}-\sqrt{15}-\sqrt{14}+\sqrt6}{4}+\frac{\sqrt{35}-\sqrt{14}-\sqrt15+\sqrt6}{3}\\ \\ E=\frac{3\sqrt{35}-3\sqrt{15}-3\sqrt{14}+3\sqrt6+4\sqrt{35}-4\sqrt{14}-4\sqrt{15}+4\sqrt6}{12}\\ \\ E=\frac{7\sqrt{35}-7\sqrt{15}-7\sqrt{14}+7\sqrt6}{12}$

Logo E é um número irracional e pertence a R+

Answer 2

$E=\left[\frac{\sqrt{5}-\sqrt{2}}{\sqrt{7}+\sqrt{3}}\right]+\left[\frac{\sqrt{7}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}\right]\\\\\\E=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{7}+\sqrt{3}}+\frac{\sqrt{7}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}\\\\\\E=\frac{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})+(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{5}+\sqrt{2})}\\\\\\E=\frac{(5-2)+(7-3)}{\sqrt{35}+\sqrt{14}+\sqrt{15}+\sqrt{6}}\\\\\\E=\frac{7}{\sqrt{35}+\sqrt{14}+\sqrt{15}+\sqrt{6}}$

Daí, $\boxed{E\in\text{R}_+}$