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 88: 108

Answer

$\frac{x-y+1}{(x-y)^{2}}; x \ne y;$

Work Step by Step

$(x-y)^{-1} + (x-y)^{-2} $ $= \frac{1}{(x-y)} + \frac{1}{(x-y)^{2}} ; x \ne y;$ $[a^{-m} = \frac{1}{a^{m}}]$ Taking LCD, $= \frac{x-y+1}{(x-y)^{2}}; x \ne y;$
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