Giải thích các bước giải:
$\to M=(\dfrac{x}{(x-3)(x+3)}-\dfrac{x-3}{x(x+3)}):\dfrac{2x-3}{x(x-3)}$
$\to M=\dfrac{x^2-(x-3)^2}{x(x-3)(x+3)}.\dfrac{x(x-3)}{2x-3}$
$\to M=\dfrac{3(2x-3)}{x+3}.\dfrac{1}{2x-3}$
$\to M=\dfrac{3}{x+3}$
$\to M\in Z\to x+3\in\{1,3,-1,-3\}\to x\in\{-2,0,-4,-6\}$