ta có 10-2n\(⋮\)n-1
\(\Rightarrow\)12-(2n-2)\(⋮\)n-1
mà 2n-2\(⋮\)n-1
\(\Rightarrow\)12\(⋮\)n-1\(\Rightarrow\)n-1\(\in\)Ư(12)={\(\pm\)1;\(\pm\)2;\(\pm\)3;\(\pm\)4;\(\pm\)6;\(\pm\)12)
n-1 | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | 5 | -5 | 6 | -6 | 12 | -12 |
n | 2 | 0 | 3 | -1 | 4 | -2 | 5 | -3 | 6 | -4 | 7 | -5 | 13 | -11 |