Đáp án:
`a.`
Ta có: `n+1 \vdots n+4`
`=> n+4-3 \vdots n+4`
`=> n+4 in Ư(3) = { +- 1;+- 3}`
`n+4=1 => n=-3`
`n+4=-1 => n=-5`
`n+4=3 => n=-1`
`n+4=-3 => n=-7`
Vậy `n in {-3;-5;-1;-7}`
$\\$
`b.`
Ta có: `2n-5 \vdots n+1`
`=> 2(n+1)-7 \vdots n+1`
`=> n+1 in Ư(7)={+- 1;+-7}`
`n+1=1 => n=0`
`n+1=-1 => n=-2`
`n+1=7 => n=6`
`n+1=-7 => n=-8`
Vậy `n in {0;-2;6;-8}`