Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 10 - Cumulative Review - Page 634: 14a

Answer

$\dfrac{ab}{b-a}$

Work Step by Step

Using $a^{-1}=\dfrac{1}{a}$, the expression $ \left( a^{-1}-b^{-1} \right)^{-1} $ simplifies to \begin{array}{l} \left( \dfrac{1}{a}-\dfrac{1}{b} \right)^{-1} \\\\= \left( \dfrac{b-a}{ab} \right)^{-1} \\\\= \dfrac{ab}{b-a} .\end{array}
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