\(n+6⋮n+2\)
\(\Leftrightarrow n+2+4⋮n+2\)
Mà \(n+2⋮n+2\)
\(\Rightarrow4⋮n+2\)
\(\Rightarrow n+2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow n\in\left\{-1;-3;0;-4;2;-6\right\}\)
\(n+6⋮n+2\)
\(\Rightarrow n+2+4⋮n+2\)
Vì \(n+2⋮n+2\)
\(\Rightarrow4⋮n+2\)
\(n+2\inƯ\left(4\right)=\left\{1;2;4\right\}\)
\(\Rightarrow n\in\left\{0;2\right\}\)
Vậy ................
à n là STN
Mình nhìn nhầm xin lỗi
Bài giải
Ta có : \(\frac{n+6}{n+2}=\frac{n+2+4}{n+2}=\frac{n+2}{n+2}+\frac{4}{n+2}=1+\frac{4}{n+2}\)
\(n+6\text{ }⋮\text{ }n+2\text{ }\Rightarrow\text{ }4\text{ }⋮\text{ }n+2\text{ }\Leftrightarrow\text{ }n+2\inƯ\left(4\right)\)
Ta có bảng :
| n + 2 | - 1 | 1 | - 2 | 2 | - 4 | 4 |
| n | 1 | 3 | 0 | 4 | - 2 | 6 |
Vậy \(n\in\left\{1\text{ ; }3\text{ ; }0\text{ ; }4\text{ ; }-2\text{ ; }6\right\}\)
Bài giải
Ta có : \(\frac{n+6}{n+2}=\frac{n+2+4}{n+2}=1+\frac{4}{n+2}\)
\(n+6\text{ }⋮\text{ }n+2\text{ }\Rightarrow\text{ }4\text{ }⋮\text{ }n+2\text{ }\Leftrightarrow\text{ }n+2\inƯ\left(4\right)\)
Ta có bảng :
| n + 2 | - 1 | 1 | - 2 | 2 | - 4 | 4 |
| n | 1 | 3 | 0 | 4 | - 2 | 6 |
Vậy \(n\in\left\{1\text{ ; }3\text{ ; }0\text{ ; }4\text{ ; }-2\text{ ; }6\right\}\)
