Answer
$-\frac{8}{(x+4)(x-4)}$.
Work Step by Step
The given expression is
$\Rightarrow \frac{\frac{x-3}{x^2-16}}{1+\frac{1}{x-4}}-\frac{1}{x-4}$
$=\frac{x-3}{x^2-16}$
Factor all terms.
$= x^2-16$
$= x^2-4^2$
Use the special formula $A^2-B^2=(A+B)(A-B)$.
$= (x+4)(x-4)$
$=\frac{x-3}{(x+4)(x-4)}$
$=1+\frac{1}{x-4}$
$=\frac{1}{1}+\frac{1}{x-4}$
The LCD of the denominators is $(x-4)$.
$=\frac{(x-4)}{(x-4)}+\frac{1}{x-4}$
$=\frac{x-4+1}{(x-4)}$
Simplify.
$=\frac{x-3}{x-4}$
Back substitute all values into the given fraction.
$\Rightarrow \frac{\frac{x-3}{(x+4)(x-4)}}{\frac{x-3}{x-4}}-\frac{1}{x-4}$
Invert the divisor and multiply.
$\Rightarrow \frac{x-3}{(x+4)(x-4)}\cdot \frac{x-4}{x-3}-\frac{1}{x-4}$
Cancel common terms.
$\Rightarrow \frac{1}{x+4}-\frac{1}{x-4}$
The LCD of the denominators is $(x+4)(x-4)$.
$\Rightarrow \frac{x-4}{(x+4)(x-4)}-\frac{x+4}{(x+4)(x-4)}$
$\Rightarrow \frac{x-4-(x+4)}{(x+4)(x-4)}$
Simplify.
$\Rightarrow \frac{x-4-x-4}{(x+4)(x-4)}$
$\Rightarrow -\frac{8}{(x+4)(x-4)}$.
