a/ Rút gọn
$\dfrac{3}{x+3}-\dfrac{6x}{x²-9}+\dfrac{x}{x+3}$
$=\dfrac{3(x-3)-6x+x(x-3)}{(x+3)(x-3)}$
$=\dfrac{3x-9-6x+x²-3x}{(x+3)(x-3)}$
$=\dfrac{x²-6x-9}{(x+3)(x-3)}$
ĐKXĐ: $\begin{cases}x+3\ne 0\\x-3\ne 0\end{cases}↔\begin{cases}x\ne -3\\x\ne 3\end{cases}$
b/ $\dfrac{x+1}{x+3}=\dfrac{x+3-2}{x+3}=1-\dfrac{2}{x+3}$
$A∈\mathbb{Z}↔x+3\vdots 2$
$↔x+3∈Ư(2)=\{±1;±2\}$
$→$ Bảng giá trị
$\begin{array}{|c|c|c|}\hline x+3&1&-1&2&-2\\\hline x&-2&-4&-1&-5\\\hline\end{array}$
Vậy $x∈\{-2;-4;-1;-5\}$