## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 10 - Exponents and Radicals - 10.5 Expressions Containing Several Radical Terms - 10.5 Exercise Set - Page 661: 68

#### Answer

$\dfrac{\sqrt{35}-\sqrt{14}+5-\sqrt{10}}{3}$

#### Work Step by Step

Factoring the $-1$ from the numerator, the given expression, $\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{5}+\sqrt{2}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{5}+\sqrt{2}}\cdot\dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\\\= \dfrac{\sqrt{7}(\sqrt{5})+\sqrt{7}(-\sqrt{2})+\sqrt{5}(\sqrt{5})+\sqrt{5}(-\sqrt{2})}{(\sqrt{5})^2-(\sqrt{2})^2} \\\\= \dfrac{\sqrt{7(5)}-\sqrt{7(2)}+\sqrt{5(5)}-\sqrt{5(2)}}{5-2} \\\\= \dfrac{\sqrt{35}-\sqrt{14}+\sqrt{(5)^2}-\sqrt{10}}{3} \\\\= \dfrac{\sqrt{35}-\sqrt{14}+5-\sqrt{10}}{3} .\end{array}

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