n+3\(⋮\)n+1
=> n+1+2\(⋮\)n+1
=> 2\(⋮\)n+1
=> n+1 \(\in\)1,2,-1,-2
=> n \(\in\)-2,1-3,-4
\(n+3⋮n+1\)
\(n+3=n+1+2⋮n+1\)
mà \(n+1⋮n+1\)
\(\Rightarrow2⋮n+1\)
\(\Rightarrow n+1\inƯ\left(2\right)\)
n+1 | 1 | 2 |
n | 0 | 1 |
Vậy \(n\in\left\{0;1\right\}\)
nếu sai thì cho mk xin lỗi nhé
\(Ta có: n+3 chia hết cho n+1 => n+3-(n+1) chia hết cho n+1 suy ra 2 chia hết cho n+1 suy ra n+1 E {+-1;+-2} suy ra n E { -2;0;-3;1 } k mk nha \)