\(\frac{2n+15}{n+1}=\frac{2n+2+13}{n+1}=\frac{2\left(n+1\right)+13}{n+1}=\frac{2\left(n+1\right)}{n+1}+\frac{13}{n+1}=2+\frac{13}{n+1}\)
Để \(\frac{2n+15}{n+1}\) là số nguyên <=> \(n+1\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
Ta có bảng:
| n + 1 | 1 | -1 | 13 | -13 |
| n | 0 | -2 | 12 | -14 |
Vậy n = {0;-2;12;-14}