Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Review - Exercises: 46

Answer

$\dfrac{1}{x+2}+\dfrac{1}{x^{2}-4}-\dfrac{2}{x^{2}-x-2}=\dfrac{x^{2}-2x-5}{(x+2)(x-2)(x+1)}$

Work Step by Step

$\dfrac{1}{x+2}+\dfrac{1}{x^{2}-4}-\dfrac{2}{x^{2}-x-2}$ Factor the denominators of the second and third terms: $\dfrac{1}{x+2}+\dfrac{1}{x^{2}-4}-\dfrac{2}{x^{2}-x-2}=...$ $...=\dfrac{1}{x+2}+\dfrac{1}{(x-2)(x+2)}-\dfrac{2}{(x-2)(x+1)}=...$ Evaluate the operations indicated using the LCD (Least common denominator) of the rational expressions and simplify if possible: $...=\dfrac{(x-2)(x+1)+(x+1)-2(x+2)}{(x+2)(x-2)(x+1)}=...$ $...=\dfrac{x^{2}-x-2+x+1-2x-4}{(x+2)(x-2)(x+1)}=\dfrac{x^{2}-2x-5}{(x+2)(x-2)(x+1)}$
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