解:$(\frac{3}{2-x}-x-2)÷\frac{x-1}{x^2-4x+4}$
$=(\frac{3}{2-x}-\frac{(x+2)(2-x)}{2-x})÷\frac{x-1}{(x-2)^2}$
$=\frac{3-(4-x^2)}{2-x}·\frac{(x-2)^2}{x-1}$
$=\frac{3-4+x^2}{-(x-2)}·\frac{(x-2)^2}{x-1}$
$=\frac{x^2-1}{-(x-2)}·\frac{(x-2)^2}{x-1}$
$=\frac{(x+1)(x-1)}{-(x-2)}·\frac{(x-2)^2}{x-1}$
$=-(x+1)(x-2)$
$=-x^2+2x-x+2$
$=-x^2+x+2$