Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Prerequisites - P.2 - Exponents and Radicals - Exercises - Page 25: 54

Answer

$3\sqrt 6-3\sqrt 5=3(\sqrt 6-\sqrt 5)$

Work Step by Step

$\frac{3}{\sqrt 5+\sqrt 6}=\frac{3}{\sqrt 6+\sqrt 5}=\frac{3}{\sqrt 6+\sqrt 5}\frac{\sqrt 6-\sqrt 5}{\sqrt 6-\sqrt 5}=\frac{3\sqrt 6-3\sqrt 5}{\sqrt 6\sqrt 6-\sqrt 6\sqrt 5+\sqrt 5\sqrt 6-\sqrt 5\sqrt 5}=\frac{3\sqrt 6-3\sqrt 5}{(\sqrt 6)^2-(\sqrt 5)}=\frac{3\sqrt 6-3\sqrt 5}{6-5}=\frac{3\sqrt 6-3\sqrt 5}{1}=3\sqrt 6-3\sqrt 5$
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