a: \(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{\left(x-2\right)\cdot\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)}{x}\)
\(=\dfrac{-4x}{x\left(x+2\right)}=\dfrac{-4}{x+2}\)
b: Ta có: 2x+x=0
=>3x=0
hay x=0(loại)
c: Để A=1/2thì \(\dfrac{-4}{x+2}=\dfrac{1}{2}\)
=>x+2=-8
hay x=-10
d: Để A là số nguyên dương thì \(\left\{{}\begin{matrix}x>-2\\x+2\in\left\{1;-1;2;-2;4;-4\right\}\end{matrix}\right.\Leftrightarrow x\in\left\{-1;0;2\right\}\)