Ответ:

[tex]\frac{1}{a-2}[/tex]

Объяснение:

[tex]\frac{1}{2a-4}+\frac{a+2}{2(2-a)^2}-\frac{2}{a^2-4a+4}=\frac{1}{2(a-2)}+\frac{a+2}{2(a-2)^2}-\frac{2}{(a-2)^2}=\\\\\frac{a-2}{2(a-2)^2}+\frac{a+2}{2(a-2)^2}-\frac{2\cdot 2}{2(a-2)^2}=\frac{a-2+a+2-4}{2(a-2)^2}=\\\\\frac{2a-4}{2(a-2)^2}=\frac{2(a-2)}{2(a-2)^2}=\frac{1}{a-2}[/tex]