Đáp án:
$a)\text{ĐKXĐ}: \left\{\begin{array}{l} x \ne -3\\ x\ne 2\end{array} \right.\\ b)A=\dfrac{x-3}{x-2}\\ c)A x \in\{1;3\}$
Giải thích các bước giải:
$A=\dfrac{x+2}{x+3}-\dfrac{5}{(x+3)(x-2)}\\ a)\text{ĐKXĐ}: \left\{\begin{array}{l} x+3 \ne 0\\ (x+3)(x-2) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne -3\\ x\ne 2\end{array} \right.\\ b)A=\dfrac{x+2}{x+3}-\dfrac{5}{(x+3)(x-2)}\\ =\dfrac{(x+2)(x-2)}{(x+3)(x-2)}-\dfrac{5}{(x+3)(x-2)}\\ =\dfrac{(x+2)(x-2)-5}{(x+3)(x-2)}\\ =\dfrac{x^2-4-5}{(x+3)(x-2)}\\ =\dfrac{x^2-9}{(x+3)(x-2)}\\ =\dfrac{(x-3)(x+3)}{(x+3)(x-2)}\\ =\dfrac{x-3}{x-2}\\ c)A=\dfrac{x-3}{x-2} \in \mathbb{Z}\\ \Leftrightarrow A=\dfrac{x-2-1}{x-2} \in \mathbb{Z}\\ \Leftrightarrow A =1-\dfrac{1}{x-2} \in \mathbb{Z}\\ \Rightarrow \dfrac{1}{x-2} \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow (x-2) \in Ư(1)\\ \Rightarrow x \in\{1;3\}$
