College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 87: 72

Answer

$\frac{1}{(x+1)(x+h+1)}; x \ne -1, x\ne-h-1, h \ne 0;$

Work Step by Step

$\frac{\frac{x+h}{x+h+1}- \frac{x}{x+1}}{h}$ Take LCD in the numerator, $=\frac{\frac{(x+h)(x+1)-(x)(x+h+1)}{(x+h+1)(x+1)}}{h}; x \ne -1, x\ne-h-1, h \ne 0;$ $=\frac{\frac{x^{2}+hx+x+h-(x^{2}+hx+x)}{(x+h+1)(x+1)}}{h}; x \ne -1, x\ne-h-1, h \ne 0;$ $=\frac{\frac{x^{2}+hx+x+h-x^{2}-hx-x}{(x+h+1)(x+1)}}{h}; x \ne -1, x\ne-h-1, h \ne 0;$ $=\frac{\frac{h}{(x+h+1)(x+1)}}{h}; x \ne -1, x\ne-h-1, h \ne 0;$ $=\frac{h}{(x+h+1)(x+1)} \times \frac{1}{h} ; x \ne -1, x\ne-h-1, h \ne 0;$ $= \frac{1}{(x+h+1)(x+1)}; x \ne -1, x\ne-h-1, h \ne 0;$
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