\(\frac{5n-3}{3-n}=\frac{5n-15}{3-n}+\frac{12}{3-n}=\frac{-5.\left(3-n\right)}{3-n}+\frac{12}{3-n}=-5+\frac{12}{3-n}\)
để \(\frac{5n-3}{3-n}\)là số nguyên thì: \(-5+\frac{12}{3-n}\in Z\)
=>3-n tuộc Ư(12)={1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}
ta có bảng sau:
| 3-n | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | 6 | -6 | 12 | -12 |
| n | 2 | 4 | 1 | 5 | 0 | 6 | -1 | 7 | -3 | 9 | -9 | 15 |
Vậy n={2;4;1;5;0;6;-1;7;-3;9;-9;15}