`A=\frac{x+2}{x+3}-\frac{5}{(x-2)(x+3)}`
đkxđ : `x\ne 2 ; -3`
`->A=\frac{(x+2)(x-2)}{(x+3)(x-2)}-\frac{5}{(x-2)(x+3)}`
`->A=\frac{x^2-4-5}{(x-2)(x+3)}`
`->A=\frac{x^2-9}{(x-2)(x+3)}`
`->A=\frac{(x-3)(x+3)}{(x-2)(x+3)}`
`->A=\frac{x-2-1}{x-2}`
`->A=1-\frac{1}{x-2}`
Để `A` nguyên thì `\frac{1}{x-2}` nguyên
ĐK : `x\ne 2`
`->x-2∈Ư_{1}`
`->x-2={+-1}`
$\left[ \begin{array}{l}x-2=1\\x-2=-1\end{array} \right.$
$\to \left[ \begin{array}{l}x=3(tm)\\x=1(tm)\end{array} \right.$
Vậy với `x=3` thì `A=1-\frac{1}{1}=0`
Với `x=1` thì `A=1-\frac{1}{-1}=2`
