## Precalculus (6th Edition) Blitzer

The simplified form of the expression $\frac{1-\frac{x}{x+2}}{1+\frac{1}{x}}$ is $\frac{2x}{{{x}^{2}}+3x+2}$.
Consider the expression: $\frac{1-\frac{x}{x+2}}{1+\frac{1}{x}}$ Multiply the numerator as well as the denominator by $\left( x+2 \right)x$ \begin{align} & \frac{1-\frac{x}{x+2}}{1+\frac{1}{x}}=\frac{\left( 1-\frac{x}{x+2} \right)\left( x+2 \right)x}{\left( 1+\frac{1}{x} \right)\left( x+2 \right)x} \\ & =\frac{\left( x+2 \right)x-\frac{x}{\left( x+2 \right)}\left( x+2 \right)x}{\left( x+2 \right)x+\frac{1}{x}\left( x+2 \right)x} \\ & =\frac{\left( x+2 \right)x-{{x}^{2}}}{\left( x+2 \right)x+x+2} \\ & =\frac{{{x}^{2}}+2x-{{x}^{2}}}{{{x}^{2}}+2x+x+2} \end{align} On further simplification, we get: \begin{align} & \frac{1-\frac{x}{x+2}}{1+\frac{1}{x}}=\frac{{{x}^{2}}+2x-{{x}^{2}}}{{{x}^{2}}+2x+x+2} \\ & =\frac{2x}{{{x}^{2}}+3x+2} \end{align} Therefore, the simplified form of the expression $\frac{1-\frac{x}{x+2}}{1+\frac{1}{x}}$ is $\frac{2x}{{{x}^{2}}+3x+2}$.