a)\(n+7⋮n+2\)
\(\Rightarrow\left(n+2\right)+5⋮n+2\)
\(\Rightarrow5⋮n+2\)
\(\Rightarrow n+2\inƯ\left(5\right)\)
Ta có:\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
| n+2 | 1 | -1 | 5 | -5 |
| n | -1 | -3 | 3 | -7 |
Vậy \(n\in\left\{-1;-3;3;7\right\}\)
b)\(9-n⋮n-3\)
\(\Rightarrow6-\left(n-3\right)⋮n-3\)
\(\Rightarrow6⋮n-3\)
\(\Rightarrow n-3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| n-3 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
| n | 4 | 2 | 5 | 1 | 6 | 0 | 9 | -3 |
Vậy\(n\in\left\{4;2;5;1;6;0;9;-3\right\}\)
c)\(n^2+n+17⋮n+1\)
\(\Rightarrow n\left(n+1\right)+17⋮n+1\)
\(\Rightarrow17⋮n+1\)
\(\Rightarrow n+1\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
| n+1 | 1 | -1 | 17 | -17 |
| n | 0 | -2 | 18 | -18 |
Vậy \(n\in\left\{0;-2;18;-18\right\}\)
e) \(2n+7⋮n+1\)
\(\Rightarrow2n+7-2\left(n+1\right)⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\inƯ\left(5\right)\)
\(\Rightarrow\left[\begin{matrix}n+1=1\Rightarrow n=0\\n+1=5\Rightarrow n=4\end{matrix}\right.\)
\(\Rightarrow n=0;n=4\)
đánh dấu thôi nha