## Precalculus (6th Edition) Blitzer

The required solution is $\frac{x-1}{x+3}$
We have the given rational expression: $\left( 1-\frac{1}{x} \right)\left( 1-\frac{1}{x+1} \right)\left( 1-\frac{1}{x+2} \right)\left( 1-\frac{1}{x+3} \right)$ Solve the first bracket of the given expression: \begin{align} & 1-\frac{1}{x}=1\times \frac{x}{x}-\frac{1}{x} \\ & =\frac{x}{x}-\frac{1}{x} \\ & =\frac{x-1}{x} \end{align} Also, solve the second bracket of the given expression: \begin{align} & 1-\frac{1}{x+1}=1\times \frac{\left( x+1 \right)}{\left( x+1 \right)}-\frac{1}{x+1} \\ & =\frac{x+1}{x+1}-\frac{1}{x+1} \\ & =\frac{x+1-1}{x+1} \\ & =\frac{x}{x+1} \end{align} Solve the third bracket of the given expression: \begin{align} & 1-\frac{1}{x+2}=1\times \frac{\left( x+2 \right)}{\left( x+2 \right)}-\frac{1}{x+2} \\ & =\frac{x+2}{x+2}-\frac{1}{x+2} \\ & =\frac{x+2-1}{x+2} \\ & =\frac{x+1}{x+2} \end{align} And solve the fourth bracket of the given expression: \begin{align} & 1-\frac{1}{x+3}=1\times \frac{\left( x+3 \right)}{\left( x+3 \right)}-\frac{1}{x+3} \\ & =\frac{x+3}{x+3}-\frac{1}{x+3} \\ & =\frac{x+3-1}{x+3} \\ & =\frac{x+2}{x+3} \end{align} Simplifying the given rational expression: \begin{align} & \left( 1-\frac{1}{x} \right)\left( 1-\frac{1}{x+1} \right)\left( 1-\frac{1}{x+2} \right)\left( 1-\frac{1}{x+3} \right)=\left( \frac{x-1}{x} \right)\left( \frac{x}{x+1} \right)\left( \frac{x+1}{x+2} \right)\left( \frac{x+2}{x+3} \right) \\ & =\frac{x-1}{x+3} \end{align} Hence, $\left( 1-\frac{1}{x} \right)\left( 1-\frac{1}{x+1} \right)\left( 1-\frac{1}{x+2} \right)\left( 1-\frac{1}{x+3} \right)=$ $\frac{x-1}{x+3}$.