## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Review - Exercises: 48

#### Answer

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

#### Work Step by Step

$\dfrac{\dfrac{1}{x}-\dfrac{1}{x+1}}{\dfrac{1}{x}+\dfrac{1}{x+1}}$ Evaluate the subtraction indicated in the numerator and the sum indicated on the denominator: $\dfrac{\dfrac{1}{x}-\dfrac{1}{x+1}}{\dfrac{1}{x}+\dfrac{1}{x+1}}=\dfrac{\dfrac{(x+1)-x}{x(x+1)}}{\dfrac{(x+1)+x}{x(x+1)}}=...$ Evaluate the division and simplify: $...=\dfrac{(x+1)-x}{x(x+1)}\div\dfrac{(x+1)+x}{x(x+1)}=\dfrac{[(x+1)-x][x(x+1)]}{[(x+1)+x][x(x+1)]}=...$ $...=\dfrac{x+1-x}{x+1+x}=\dfrac{1}{2x+1}$

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