có 2n+1\(⋮\)2n+1=>6n+3\(⋮\)2n+1
mà -3n+2\(⋮\)2n+1=>6n-4\(⋮\)2n+1
=>(6n+3)-(6n-4)\(⋮\)2n+1
=>6n+3-6n+4\(⋮\)2n+1
7\(⋮\)2n+1
2n+1\(\inƯ\left(7\right)\)
=>2n+1\(\in\left\{\pm1;\pm7\right\}\)
Ta có bảng sau:
2n+1 | 1 | -1 | 7 | -7 |
2n | 0 | -2 | 6 | -8 |
n | 0 | -1 | 3 | -4 |
Vậy n\(\in\left\{0;-1;3;-4\right\}\)