Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises: 88

Answer

$\dfrac{1}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{x-1}$

Work Step by Step

$\dfrac{1}{\sqrt{x}+1}$ Multiply both numerator and denominator by $-\sqrt{x}+1$ and simplify: $\dfrac{1}{\sqrt{x}+1}=\Big(\dfrac{1}{\sqrt{x}+1}\Big)\Big(\dfrac{-\sqrt{x}+1}{-\sqrt{x}+1}\Big)=\dfrac{1-\sqrt{x}}{1-x}=\dfrac{\sqrt{x}-1}{x-1}$
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