College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.5 - Rational Expressions - R.5 Exercises - Page 48: 77

Answer

$\dfrac{-2x-h}{[(x+h)^2+9](x^2+9)}$

Work Step by Step

The given expression, $ \dfrac{\dfrac{1}{(x+h)^2+9}-\dfrac{1}{x^2+9}}{h} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{\dfrac{(x^2+9)(1)-[(x+h)^2+9](1)}{[(x+h)^2+9](x^2+9)}}{h} \\\\= \dfrac{(x^2+9)(1)-[(x+h)^2+9](1)}{h[(x+h)^2+9](x^2+9)} \\\\= \dfrac{x^2+9-(x+h)^2-9}{h[(x+h)^2+9](x^2+9)} \\\\= \dfrac{x^2-(x+h)^2}{h[(x+h)^2+9](x^2+9)} \\\\= \dfrac{x^2-(x^2+2xh+h^2)}{h[(x+h)^2+9](x^2+9)} \\\\= \dfrac{x^2-x^2-2xh-h^2}{h[(x+h)^2+9](x^2+9)} \\\\= \dfrac{-2xh-h^2}{h[(x+h)^2+9](x^2+9)} \\\\= \dfrac{h(-2x-h)}{h[(x+h)^2+9](x^2+9)} \\\\= \dfrac{\cancel{h}(-2x-h)}{\cancel{h}[(x+h)^2+9](x^2+9)} \\\\= \dfrac{-2x-h}{[(x+h)^2+9](x^2+9)} .\end{array}
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