\(A=\dfrac{n^2+3n+3}{n-1}\left(n\ne1\right)\\ =\dfrac{n^2-2n+1+5n+2}{n-1}\\ =\dfrac{\left(n-1\right)^2+5n-5+7}{n-1}\\ =\dfrac{\left(n-1\right)^2}{n-1}+\dfrac{5n-5}{n-1}+\dfrac{7}{n-1}\\ =n-1+5+\dfrac{7}{n-1}\)
Để A là số nguyên thì \(n-1\in Z\) và \(\dfrac{7}{n-1}\in Z \Rightarrow n-1\inƯ\left(7\right)\)
n-1 | n |
-7 | -6 |
-1 | 0 |
1 | 2 |
7 | 8 |
Vậy \(n\in\left\{-6;0;2;8\right\}\)