Answer
$\frac{x-1}{x+3}$.
Work Step by Step
The given expression is
$\Rightarrow \left (1-\frac{1}{x} \right )\left (1-\frac{1}{x+1} \right )\left (1-\frac{1}{x+2} \right )\left (1-\frac{1}{x+3} \right )$
Multiply the numerators and the denominators to equal all denominators.
$\Rightarrow \left (\frac{x}{x}-\frac{1}{x} \right )\left (\frac{x+1}{x+1}-\frac{1}{x+1} \right )\left (\frac{x+2}{x+2}-\frac{1}{x+2} \right )\left (\frac{x+3}{x+3}-\frac{1}{x+3} \right )$
Add all the numerators in the bracket because denominators are equal.
$\Rightarrow \left (\frac{x-1}{x} \right )\left (\frac{x+1-1}{x+1} \right )\left (\frac{x+2-1}{x+2} \right )\left (\frac{x+3-1}{x+3} \right )$
Simplify.
$\Rightarrow \left (\frac{x-1}{x} \right )\left (\frac{x}{x+1} \right )\left (\frac{x+1}{x+2} \right )\left (\frac{x+2}{x+3} \right )$
Cancel common terms.
$\Rightarrow \frac{x-1}{x+3}$.
