Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises - Page 76: 104

Answer

$\frac{3\sqrt 5-2\sqrt 3+3\sqrt {15}-6}{33}$

Work Step by Step

Multiply $\frac{3\sqrt 5-2\sqrt 3}{3\sqrt 5-2\sqrt 3}$ to the expression, we have: $\frac{1+\sqrt 3}{3\sqrt 5+2\sqrt 3}\cdot \frac{3\sqrt 5-2\sqrt 3}{3\sqrt 5-2\sqrt 3}=\frac{3\sqrt 5-2\sqrt 3+3\sqrt {15}-6}{45-12}=\frac{3\sqrt 5-2\sqrt 3+3\sqrt {15}-6}{33}$
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