n+3 \(⋮\)n-1
=> (n-1)+4 \(⋮\)n-1
Vì n-1 \(⋮\)n-1 nên 4 \(⋮\)n-1
=> n-1 \(\in\) Ư(4) = {1;2;4}
n-1 | 1 | 2 | 4 |
n | 2 | 3 | 5 |
n+3 \(⋮\) n-1
Ta có : n-1\(⋮\) n-1
Theo bài ra : n+3 \(⋮\)n-1
n+3-(n-1) \(⋮\)n-1
n+3-n+1 \(⋮\) n-1
4 \(⋮\)n-1
\(\Rightarrow\) n-1 \(\in\) Ư(4) = { 1; 2; 4 }
n-1=1 n-1=2 n-1 =4
n=2 n=3 n=5
Vậy n \(\in\) {2;3;5}