Tìm `x` thuộc `Z` để `A` thuộc `Z` `A=(x+2)/(x+3)-5/((x+3)(x-2))`

Đáp án:

$x \in \{1;3\}$

Giải thích các bước giải:

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