Ta có: \(\dfrac{n+3}{n+1}=\dfrac{n+1+2}{n+1}=1+\dfrac{2}{n+1}\)
Để \(n+3⋮n+1\) thì \(2⋮n+1\) \(\Rightarrow n+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Ta có bảng sau:
\(n+1\) | -2 | -1 | 1 | 2 |
\(n\) | -3 | -2 | 0 | 1 |
Vậy để \(n+3⋮n+1\Leftrightarrow n\in\left\{-3;-2;0;1\right\}\)
n+3 \(⋮\) n+1
- Ta có : n+3 = n+1+2 = (n+1)+2
Để (n+3) \(⋮\) (n+1) hay (n+1)+2 \(⋮\) (n+1) mà n+1 \(⋮\) (n+1) => 2\(⋮\) n-1 => n-1 E Ư(2) = {1;2}
=> n E {2;4}
Vậy n E {2;4}