Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Appendix A - Appendix A Exercises: 20


$-\dfrac {12}{x^{2}-9}$

Work Step by Step

$\dfrac {2}{x+3}-\dfrac {2}{x-3}=\dfrac {2\left( x-3\right) -2\left( x+3\right) }{\left( x+3\right) \left( x-3\right) }=\dfrac {2x-6-2x-6}{x^{2}-9}=-\dfrac {12}{x^{2}-9}$
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