Answer
The required solution is $\frac{x-1}{x+3}$
Work Step by Step
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, $\left( 1-\frac{1}{x} \right)\left( 1-\frac{1}{x+1} \right)\left( 1-\frac{1}{x+2} \right)\left( 1-\frac{1}{x+3} \right)=$ $\frac{x-1}{x+3}$.
