Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.7 Radicals - R.7 Exercises - Page R-29: 34

Answer

$\dfrac{5-\sqrt{10}}{3}$

Work Step by Step

Rationalize the denominator by multiplying $\sqrt5-\sqrt{2}$ to both the numerator and the denominator to have: $=\dfrac{\sqrt5}{\sqrt5+\sqrt{2}} \cdot \dfrac{\sqrt5-\sqrt{2}}{\sqrt5-\sqrt{2}} \\=\dfrac{5-\sqrt{10}}{5-2} \\=\dfrac{5-\sqrt{10}}{3}$
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