`2x + 5 \vdots x + 1`
`=> (2x + 5) - [2 . (x + 1)] \vdots x + 1`
`=> (2x + 5) - (2x + 2) \vdots x + 1`
`=> 2x + 5 - 2x - 2 \vdots x + 1`
`=> 5 - 2 \vdots x + 1`
`=> 3 \vdots x + 1`
`=> x + 1 ∈ Ư(3)`
`=> x + 1 ∈ {-1 ; 1 ; -3 ; 3}`
`=> x ∈ {-2 ; 0 ; -4 ; 2}`
Vậy `x ∈ {-2 ; 0 ; -4 ; 2}` thì `2x + 5 \vdots x + 1`