Để \(\frac{n+7}{n+2}\in Z\)
\(\Leftrightarrow\frac{x+2+5}{n+2}=\frac{n+2}{n+2}+\frac{5}{n+2}=1+\frac{5}{n+2}\in Z\)
Mà \(1\in Z\Leftrightarrow\frac{5}{n+2}\in Z\)
\(\Leftrightarrow n+2\inƯ_5=\left\{1;5;-1;-5\right\}\)
\(\Leftrightarrow n=\left\{-1;3;-3;-7\right\}\)
Mà \(n\in N\Rightarrow n=\left\{3\right\}\)
\(n+7⋮n+2\)
\(n+2+5⋮n+2\)
Mà \(n+2⋮n+2\Rightarrow5⋮n+2\)hay \(n+2\inƯ\left(5\right)=\left\{+-1;+-5\right\}\)
Ta có bảng :
| n+2 | 1 | -1 | 5 | -5 |
| n | -1 | -3 | 3 | -7 |
Mà n là số tự nhiên => n = 5
Vậy, n = 5
\(\frac{n+7}{n+2}=\frac{n+2+5}{n+2}=\frac{5}{n+2}\)
hay \(x+2\inƯ\left(5\right)=\left\{1;5\right\}\)
| x + 2 | 1 | 5 |
| x | -1 | 3 |
