B=\(\dfrac{2.n+1+3.n+5-4.n+5}{n-3}\)
B=\(\dfrac{5.n+6-4.n+5}{n-3}\)
B=\(\dfrac{n+1}{n-3}\)
B=\(\dfrac{n-3+4}{n-3}\)
B=\(\dfrac{n-3}{n-3}\)+\(\dfrac{4}{n-3}\)
B=1+\(\dfrac{4}{n-3}\)
Để B nguyên thì 4\(⋮\)n-3 hay n-3\(\in\)Ư(4).Ta có bảng sau:
n-3 | 1 | 2 | 4 | -1 | -2 | -4 |
n | 4 | 5 | 7 | 2 | 1 |
-1 |
Vậy n\(\in\){ 4;5;7;2;1;-1)