$\\$
`3n+4 \vdots n-5`
`-> 3n - 15 + 19 \vdots n-5`
`->3 (n-5) + 19 \vdots n-5`
Vì `n-5 \vdots n-5 -> 3 (n-5) \vdots n-5`
`-> 19 \vdots n-5`
`->n-5 ∈ Ư (19)={1;-1;19;-19}`
$\bullet$ `n-5=1 -> n=6` (tm)
$\bullet$ `n-5=-1 ->n=4` (tm)
$\bullet$ `n-5=19 ->n=24` (tm)
$\bullet$ `n-5=-19 ->n=-14` (ktm)
Vậy `n ∈ {6; 4; 24}` để `3n+4 \vdots n-5`