Answer
$\Rightarrow -\frac{4}{(x+2)(x-2)}$.
Work Step by Step
The given expression is
$\Rightarrow \frac{\frac{x-1}{x^2-4}}{1+\frac{1}{x-2}}-\frac{1}{x-2}$
Solve the numerator of the first fraction.
$=\frac{x-1}{x^2-4}$
Factor all terms.
$= x^2-4$
$= x^2-2^2$
Use the special formula $A^2-B^2=(A+B)(A-B)$.
$= (x+2)(x-2)$
$=\frac{x-1}{(x+2)(x-2)}$
Solve the denominator of the first fraction.
$=1+\frac{1}{x-2}$
$=\frac{1}{1}+\frac{1}{x-2}$
The LCD of the denominators is $(x-2)$.
$=\frac{(x-2)}{(x-2)}+\frac{1}{x-2}$
$=\frac{x-2+1}{(x-2)}$
Simplify.
$=\frac{x-1}{x-2}$
Back substitute all values into the given fraction.
$\Rightarrow \frac{\frac{x-1}{(x+2)(x-2)}}{\frac{x-1}{x-2}}-\frac{1}{x-2}$
Invert the divisor and multiply.
$\Rightarrow \frac{x-1}{(x+2)(x-2)}\cdot \frac{x-2}{x-1}-\frac{1}{x-2}$
Cancel common terms.
$\Rightarrow \frac{1}{x+2}-\frac{1}{x-2}$
The LCD of the denominators is $(x+2)(x-2)$.
$\Rightarrow \frac{x-2}{(x+2)(x-2)}-\frac{x+2}{(x+2)(x-2)}$
$\Rightarrow \frac{x-2-(x+2)}{(x+2)(x-2)}$
Simplify.
$\Rightarrow \frac{x-2-x-2}{(x+2)(x-2)}$
$\Rightarrow -\frac{4}{(x+2)(x-2)}$.
