## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Review - Exercises: 45

#### Answer

$\dfrac{2}{x}+\dfrac{1}{x-2}+\dfrac{3}{(x-2)^{2}}=\dfrac{3x^{2}-7x+8}{x(x-2)^{2}}$

#### Work Step by Step

$\dfrac{2}{x}+\dfrac{1}{x-2}+\dfrac{3}{(x-2)^{2}}$ Evaluate the sum of the three rational expressions using their LCD (Least common denominator) and simplify if possible: $\dfrac{2}{x}+\dfrac{1}{x-2}+\dfrac{3}{(x-2)^{2}}=\dfrac{2(x-2)^{2}+x(x-2)+3x}{x(x-2)^{2}}=...$ $...=\dfrac{2(x^{2}-4x+4)+x^{2}-2x+3x}{x(x-2)^{2}}=...$ $...=\dfrac{2x^{2}-8x+8+x^{2}-2x+3x}{x(x-2)^{2}}=\dfrac{3x^{2}-7x+8}{x(x-2)^{2}}$

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