## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.6 - Rational Expressions - Exercise Set - Page 87: 119

#### Answer

The required solution is $\frac{x-y+1}{{{\left( x-y \right)}^{2}}}$

#### Work Step by Step

We have the given rational expression: ${{\left( x-y \right)}^{-1}}+{{\left( x-y \right)}^{-2}}$ Solve the first bracket of the given expression: \begin{align} & {{\left( x-y \right)}^{-1}}=\frac{1}{{{\left( x-y \right)}^{1}}} \\ & =\frac{1}{x-y} \end{align} Also, solve the second bracket of the given expression: \begin{align} & {{\left( x-y \right)}^{-2}}=\frac{1}{{{\left( x-y \right)}^{2}}} \\ & =\frac{1}{\left( x-y \right)\left( x-y \right)} \end{align} And simplify the given rational expression: \begin{align} & {{\left( x-y \right)}^{-1}}+{{\left( x-y \right)}^{-2}}=\frac{1}{x-y}+\frac{1}{\left( x-y \right)\left( x-y \right)} \\ & =\frac{1}{x-y}\times \frac{\left( x-y \right)}{\left( x-y \right)}+\frac{1}{\left( x-y \right)\left( x-y \right)} \\ & =\frac{x-y}{\left( x-y \right)\left( x-y \right)}+\frac{1}{\left( x-y \right)\left( x-y \right)} \\ & =\frac{x-y+1}{\left( x-y \right)\left( x-y \right)} \end{align} Hence, ${{\left( x-y \right)}^{-1}}+{{\left( x-y \right)}^{-2}}=$ $\frac{x-y+1}{{{\left( x-y \right)}^{2}}}$.

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