Giải thích các bước giải:
a.Để $A$ xác định
$\to\begin{cases}x^2-4\ne 0\\ x-2\ne 0\\ x+2\ne 0\end{cases}$
$\to\begin{cases}(x-2)(x+2)\ne 0\\ x\ne 2\\ x\ne -2\end{cases}$
$\to x\ne \pm2$
b.Ta có:
$A=\dfrac{x^2+4}{x^2-4}+\dfrac{2}{x-2}-\dfrac{x}{x+2}$
$\to A=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}+\dfrac{2}{x-2}-\dfrac{x}{x+2}$
$\to A=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}$
$\to A=\dfrac{x^2+4+2\left(x+2\right)-x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}$
$\to A=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}$
$\to A=\dfrac{4}{x-2}$
c.Để $A\in Z$
$\to \dfrac{4}{x-2}\in Z$
$\to 4\quad\vdots\quad x-2$
Mà $x\in Z\to x-2\in Z$
$\to x-2\in U(4)$
$\to x-2\in\{1,2,4,-1,-2,-4\}$
$\to x\in\{3,4,6,1,0,-2\}$
Mà $x\ne\pm2$
$\to x\in\{3,4,6,1,0\}$
