Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.7 Radicals - R.7 Exercises: 39

Answer

$\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}=-2x-2\sqrt{x^{2}+x}-1$

Work Step by Step

$\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}$ Multiply the numerator and the denominator of the given expression by $\sqrt{x}+\sqrt{x+1}$, which is the conjugate of the denominator: $\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}\cdot\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}+\sqrt{x+1}}=...$ Evaluate the products: $...=\dfrac{(\sqrt{x}+\sqrt{x+1})^{2}}{(\sqrt{x})^{2}-(\sqrt{x+1})^{2}}=...$ $...=\dfrac{(\sqrt{x})^{2}+2(\sqrt{x})(\sqrt{x+1})+(\sqrt{x+1})^{2}}{x-x-1}=...$ $...=\dfrac{x+2\sqrt{x^{2}+x}+x+1}{-1}=-2x-2\sqrt{x^{2}+x}-1$
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