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 - Page 43: 61

Answer

$\dfrac{1+\dfrac{1}{x+2}}{1-\dfrac{1}{x+2}}=\dfrac{x+3}{x+1}$

Work Step by Step

$\dfrac{1+\dfrac{1}{x+2}}{1-\dfrac{1}{x+2}}$ Evaluate the sum present in the numerator and substraction present in the denominator: $\dfrac{1+\dfrac{1}{x+2}}{1-\dfrac{1}{x+2}}=\dfrac{\dfrac{(x+2)+1}{x+2}}{\dfrac{(x+2)-1}{x+2}}=...$ Evaluate the division: $...=\dfrac{[(x+2)+1](x+2)}{[(x+2)-1](x+2)}=...$ Simplify: $...=\dfrac{x+2+1}{x+2-1}=\dfrac{x+3}{x+1}$
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