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: 43

Answer

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

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 numerator: $\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}\cdot\dfrac{\sqrt{x}-\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}=...$ Evaluate the products and simplify: $...=\dfrac{(\sqrt{x})^{2}-(\sqrt{x+1})^{2}}{(\sqrt{x}-\sqrt{x+1})^{2}}=...$ $...=\dfrac{x-x-1}{(\sqrt{x})^{2}-2(\sqrt{x})(\sqrt{x+1})+(\sqrt{x+1})^{2}}=...$ $...=-\dfrac{1}{x-2\sqrt{x^{2}+x}+x+1}=-\dfrac{1}{2x-2\sqrt{x^{2}+x}+1}$
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