a)

A= ($\frac{3}{x-1}$ - $\frac{2}{x+1}$) : $\frac{x+2}{x^2-1}$ 

  = [$\frac{3.(x+1)}{(x-1)(x+1)}$ - $\frac{2.(x-1)}{(x-1)(x+1)}$] : $\frac{x+2}{x^2-1}$ 

  = [$\frac{3x+3-2x+2}{(x-1)(x+1)}$] : $\frac{x+2}{(x-1)(x+1)}$ 

  = $\frac{x+5}{(x-1)(x+1)}$ . $\frac{(x-1)(x+1)}{x+2}$

  = $\frac{(x+5).(x-1)(x+1)}{(x-1)(x+1)(x+2)}$ 

  = $\frac{x+5}{x+2}$ 

b)

A = $\frac{x+5}{x+2}$

   = $\frac{x+2+3}{x+2}$

   = $\frac{x+2}{x+2}$ + $\frac{3}{x+2}$

   = 1 + $\frac{3}{x+2}$ 

Để A ∈ Z thì 1 + $\frac{3}{x+2}$  ∈ Z

Mà 1 ∈ Z nên $\frac{3}{x+2}$ ∈ Z

Mà x ∈ Z nên (x+2) ∈ ước (3) = {±1; ±3}

Ta có bảng sau: