Answer
$\frac{x-2}{(x+1)}; x \ne 2,3, -1;$
Work Step by Step
$\frac{x-3}{x-\frac{3}{x-2}}$
$=\frac{x-3}{\frac{x(x-2)-3}{x-2}}$
$=\frac{x-3}{\frac{x^{2}-2x-3}{x-2}}$
Factors of $x^{2}-2x-3$ are $(x-3)$ and $(x+1)$
$=\frac{x-3}{\frac{(x-3)(x+1)}{(x-2)}} ; x \ne 2,3, -1;$
$ = \frac{(x-3)(x-2)}{(x-3)(x+1)} ; x \ne 2,3, -1;$
$= \frac{(x-2)}{(x+1)}; x \ne 2,3, -1;$
