Bài 5:
a: \(A=\dfrac{x^2+2x-x^2+4x-4+12-10x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-4x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4}{x+2}\)
b: Để A là số nguyên thì \(x+2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{-1;-3;0;-4;-6\right\}\)

