a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2}=\dfrac{-3}{x-2}\)
b: Thay x=-4 vào A, ta được:
\(A=\dfrac{-3}{-4-2}=\dfrac{-3}{-6}=\dfrac{1}{2}\)
c: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)