A = \(\dfrac{3n+2}{n+1}\) (đk n \(\in\)Z ; n \(\ne\) -1)
A \(\in\) Z ⇔ 3n + 2 ⋮ n + 1
3n + 3 - 1 ⋮ n + 1
3(n +1) - 1 ⋮ n + 1
1 ⋮ n + 1
n + 1 \(\in\) { -1; 1}
n \(\in\) { -2; 0}
\(A=\dfrac{3n+2}{n+1}=\dfrac{3n+3-2}{n+1}=\dfrac{3\left(n+1\right)-2}{n+1}=3-\dfrac{2}{n+1}\)
Để A có giá trị nguyên ⇒ n+1 là Ư(2)={-1;1;-2;2}
⇒ n+1 ϵ {-1;1;-2;2}
⇒ n ϵ {-2;0;-3;1}