\(n-2⋮n+1\)
\(=>-2+n⋮1+n\)
\(=>-3+\left(1+n\right)⋮1+n\)
Do \(1+n⋮1+n\)
\(=>n+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(=>n\in\left\{-4;-2;0;2\right\}\)
\(n-2⋮n+1\)
\(n+1-3⋮n+1\)
Vì \(n+1⋮n+1\)
\(\Rightarrow n+1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
Ta cs bảng
| n+1 | 1 | -1 | 3 | -3 |
| n | 0 | -2 | 2 | -4 |
\(2n-3⋮n+1\)
\(2\left(n+1\right)-5⋮n+1\)
Vì \(2\left(n+1\right)⋮n+1\)
\(-5⋮n+1\)
\(\Rightarrow n+1\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}\)
Ta cs bảng
| n+1 | 1 | -1 | 5 | -5 |
| n | 0 | -2 | 4 | -6 |
Câu hỏi của linh tran - Toán lớp 6 - Học toán với OnlineMath
\(n-2⋮n+1\)
\(n-2=n+1-3\)
\(\Rightarrow n-2⋮n+1\Leftrightarrow n+1-3⋮n+1\Leftrightarrow-3⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(-3\right)=\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow n\in\left\{0;-2;2;-4\right\}\)
....
\(2n-3⋮n+1\)
\(2n-3=2n+2-5=2\left(n+1\right)-5\)
\(\Rightarrow2n-3⋮n+1\Leftrightarrow2\left(n+1\right)-5⋮n+1\Leftrightarrow-5⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(-5\right)=\left\{1;-1;5;-5\right\}\)
\(\Leftrightarrow n\in\left\{0;-2;4;-6\right\}\)
...
\(3n+5⋮2n-1\Leftrightarrow2\left(3n+5\right)⋮2n-1\)
\(2\left(3n+5\right)=6n+10=3\left(2n-1\right)+13\)
\(\Rightarrow2\left(3n+5\right)⋮2n-1\Leftrightarrow3\left(2n-1\right)+13⋮2n-1\Leftrightarrow13⋮2n-1\)
\(\Leftrightarrow2n-1\inƯ\left(13\right)=\left\{1;-1;13;-13\right\}\)
\(\Leftrightarrow n\in\left\{1;0;7;-6\right\}\)
....
\(2n-3⋮n+1\)
\(=>2n+2-5⋮n+1\)
Do\(2n+2⋮n+1\)
\(=>n+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(=>n\in\left\{-6;-2;0;4\right\}\)
