Giúp mình 1 bài nha help me please

`a)`

     `5 \vdots x`

`⇒ x ∈Ư(5) = {±1;±5}`

`⇒ x ∈{±1;±5}`.

`b)`

     `4 \vdots x+1`

`⇒ x+1 ∈Ư(4) = {±1;±2;±4}`

`⇒ x∈{ 0;-2;1;-3;3;-5}`.

`c)`

     `x+3 \vdots x+1`

`⇒ (x+3)/(x+1) = ( (x+1)+2 )/(x+1) = 1 + 2/(x+1)`

`⇒ 2 \vdots x+1`

`⇒ x+1 ∈Ư(2) = {±1;±2}`

`⇒ x ∈{0;-2;1;-3}`.

`d)`

      `2x+8 \vdots x+3`

`⇒ (2x+8)/(x+3) = ( (2x+6)+2 )/(x+3) = 2 + 2/(x+3)`

`⇒ 2 \vdots x+3`

`⇒ x+3 ∈Ư(2) ={±1;±2}`

`⇒ x∈{-2;-4;-1;-5}`.

`e)`

      `5x+6 \vdotsx+1`

`⇒ (5x+6)/(x+1) = ( (5x+5)+1 )/(x+1) = 5 + 1/(x+1)`

`⇒ 1 \vdots x+1`

`⇒ x+1 ∈Ư(1)={1;-1}`

`⇒ x ∈{0;-2}`.

`g)`

     `x^2 +x+3 \vdots x+1`

`⇒ (x^2 +x+3)/(x+1) = (x.x+x+3)/(x+1) = ( x.(x+1)+3 )/(x+1) = x + 3/(x+1)`

`⇒ 3 \vdots x+1`

`⇒ x+1 ∈Ư(3) ={±1;±3}`

`⇒ x∈{0;-2;2;-4}`.

`f)`

      `6x+11 \vdots 2x+3`

`⇒ (6x+11)/(2x+3) = ( (6x+9)+2 )/(2x+3) = 3 + 2/(2x+3)`

`⇒ 2 \vdots 2x+3`

`⇒ 2x+3 ∈Ư(2) = {±1;±2}`

`⇒ 2x ∈{-2;-4;-1;-5}`

`⇒ x∈{-1;-2;-1/2;-5/2}`

`Vì` `x∈ZZ ⇒ x∈{-1;-2}`.

$#nhatminhbuithien2010$