Đáp án `+` Giải thích các bước giải:

 `a, n^3+ n^2+7 vdots n+2`

`=n^2 (n+2)-(n^2+2n)+(2n+4)+3`

`=n^2 (n+2)-n(n+2)+2(n+2)+3`

`=(n+2)(n^2- n+2)+3`

Để `n^3+ n^2+7 vdots n+2`

`=> 3 vdots n+2`

`=> n+2 in Ư(3)={1,3,-1,-3}`

`=> n in {-1,1,-3,-5}`

Vậy `n in {-1,1,-3,-5}`

`b, 2n-3 vdots n^2+3`

`=> (2n-3)(2n+3) vdots n^2+ 3`

`=4n^2-9 `

`=4(n^2+3)-21`

Để `4(n^2+ 3) vdots n^2+3` thì `21 vdots n^2+3`

`=> n^2+ 3 in Ư(21)={±1,±21,±3,±7}`

`=> n^2 in {-2,-4,-24,18,0,-6,4,-10}`

`=> n^2 in {0,4}`

`=> n in {0,-2,2}`

Vậy `n in {0,-2,2}`

`c,n^5+1 vdots n^3+1`

`=n^2 (n^3+ 1)-(n^2 -1) `

`=> n^2- 1 vdots n^3+ 1`

`(n-1)(n+1) vdots (n+1)(n^2- n +1)` `(n ne -1)`

`=> n-1 vdots n^2- n+1`

`TH1: n-1=0`

`=> n=1`

`=> (1-1)(1+1) vdots (1+1)(1^2- 1+1)` `(thõa    mãn)`

`TH2: |n-1| >= n^2- n+1`

`+ n >= 1 => n-1 >= n^2- n+1`

`<=> n^2- 2n+2 <= 0`

`<=> (n-1)^2+ 1 <= 0` `(` loại `)`

`+ n<1`

`<=> n-1 >= n^2- n+1`

`<=> n^2 <=.0`

`=> n=0`

Vậy `n in {0,1}`