\(n-7⋮n+1\)
Mà \(n+1⋮n+1\)
\(\Leftrightarrow8⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(8\right)\)
Ta có bảng :
\(n+1\) | \(1\) | \(2\) | \(4\) | \(8\) | \(-1\) | \(-2\) | \(-4\) | \(-8\) |
\(n\) | \(0\) | \(1\) | \(3\) | \(7\) | \(-2\) | \(-3\) | \(-5\) | \(-9\) |
Đk \(n\in Z\) | tm | tm | tm | tm | tm | tm | tm | tm |
Vậy ...
Ta có \(n-7⋮n+1\Leftrightarrow\left(n+1\right)-8⋮\left(n+1\right)\)
Mà \(\left(n+1\right)⋮\left(n+1\right)\)
Suy ra \(8⋮\left(n+1\right)\Rightarrow n+1\inƯ_{\left(8\right)}=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
Ta có bảng
n+1 | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
n | -9 | -5 | -3 | -2 | 0 | 1 | 3 | 7 |
Mà \(n\in Z\)
Vậy \(n\in\left\{-9;-5;-3;-2;0;1;3;7\right\}\)