Để `n ∈ Z`
`=>` `(n+8)/(2n-6)` `∈ Z`
`=>` `n+8` chia hết cho `2n-6`
`=>` `2(n+8)` chia hết cho `2n-6`
`=>` `2n+16` chia hết cho `2n-6`
Ta có: `2n+16 = 2n - 6 + 22`
`=>` `22` chia hết cho `2n-6`
`=>` `2n-6` `∈` `Ư(22)` `=` `{` `±1; ±2; ±11; ±22` `}`
`=>` `2n` `∈` `{` `7; 5; 8; 4; 17; -5; 28; -16` `}`
`=>` `n` `∈` `{` `4; 2; 14; -8` `}`
Vậy `n` `∈` `{` `4; 2; 14; -8` `}`
