\(\frac{n+3}{n+1}=\frac{n+1+2}{n+1}=\frac{n+1}{n+1}+\frac{2}{n+1}=1+\frac{2}{n+1}\in Z\)
\(\Rightarrow2⋮n+1\)
\(\Rightarrow n+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow n\in\left\{0;1\right\}\left(n\in N\right)\)
(n + 3) \(⋮\) (n + 1)
\(\Rightarrow\) (n + 1 +2) \(⋮\) (n + 1)
\(\Rightarrow\) 2 \(⋮\) (n + 1)
\(\Rightarrow\) n + 1 \(\in\) Ư(2) = {1; 2}
Do n \(\in\) N, nên ta có bảng sau:
| n + 1 | 1 | 2 |
| n | 0 | 1 |
Vậy: n \(\in\) {0; 1}
Ta có : \(n+3⋮n+1\) ; Mà : \(n+1⋮n+1\)
\(\Rightarrow\left(n+3\right)-\left(n+1\right)⋮n+1\Rightarrow2⋮n+1\)
\(\Rightarrow n+1\inƯ\left(2\right)\Leftrightarrow n+1\in\left\{1;2\right\}\)
+) Nếu : \(n+1=1\Rightarrow n=1-1=0\)
+) Nếu : \(n+1=2\Rightarrow n=2-1=1\)
Vậy : \(n\in\left\{0;1\right\}\)
