Đáp án:
$x \in \{1;3\}$
Giải thích các bước giải:
$\begin{array}{l}\quad A = \dfrac{x+2}{x+3} -\dfrac{5}{(x+3)(x-2)}\qquad (x\ne -3;\, x \ne 2)\\ \to A = \dfrac{(x+2)(x-2) - 5}{(x+3)(x-2)}\\ \to A = \dfrac{x^2 - 4 - 5}{(x+3)(x-2)}\\ \to A = \dfrac{x^2 - 9}{(x+3)(x-2)}\\ \to A = \dfrac{(x+3)(x-3)}{(x+3)(x-2)}\\ \to A = \dfrac{x-3}{x-2}\\ \to A = \dfrac{x-2 - 1}{x-2}\\ \to A = 1 - \dfrac{1}{x-2}\\ \quad A \in \Bbb Z \Leftrightarrow \dfrac{1}{x-2}\in \Bbb Z\\ Do\,\,x \in \Bbb Z\\ nên\,\,x - 2 \in \Bbb Z\\ \to \dfrac{1}{x-2}\in \Bbb Z \Leftrightarrow x - 2 \in Ư(1) = \{-1;1\}\\ \to x \in \{1;3\} \end{array}$
