$Q=\dfrac{a+3}{a-2}=\dfrac{a-2+5}{a-2}=1+\dfrac{5}{a-2}$
Để Q nguyên $→$ $\dfrac{5}{a-2}$ nguyên
$→a-2\in U_{\{5\}}=\{±1;±5\}$
$→a\in\{1;3;7;-3\}$
Gọi d là UCLN ( a + 3; a- 2)
$→5\quad\vdots\quad d$
$→d=5$
$→a-2=5k→a=5k+2$
$→$ Để Q tối giản $→a\ne 5k+2$