$D =\left(\dfrac{2}{x-2} -\dfrac{4}{4-4x+4}\right): \left(\dfrac{1}{x^2 -4} +\dfrac{1}{2-x}\right)$
a) $ĐKXĐ: \begin{cases}x-2\ne 0\\4 - 4x+ 4 \ne 0\\x^2 - 4 \ne 0\\2 - x \ne 0\end{cases}$
$\Leftrightarrow (x-2)(x+2)\ne 0$
$\Leftrightarrow x \ne \pm 2$
b) $D = \left(\dfrac{2}{x-2} -\dfrac{1}{2-x}\right): \left(\dfrac{1}{(x-2)(x+2)} -\dfrac{1}{x-2}\right)$
$\to D =\dfrac{3}{x-2}:\dfrac{1 - (x+2)}{(x-2)(x+2)}$
$\to D =\dfrac{3}{x-2}\cdot\dfrac{(x-2)(x+2)}{-1-x}$
$\to D = -\dfrac{3(x+2)}{x+1}$