Đáp án:
1) Ta có: `(n+8)/(n+3)=(n+3)/(n+3)+5/(n+3)`
Vì `n+3\vdots n+3`
`⇒ 5\vdots n+3`
`⇒ n+3\in Ư(5)=\{±1;±5\}`
`· n+3=-1⇒n=-4`
`· n+3=1⇒n=-2`
`· n+3=-5⇒n=-8`
`· n+3=5⇒n=2`
Vậy `n\in \{-4;±2;-8\}`
2) Ta có: `(n+9)/(n+2)=(n+2)/(n+2)+7/(n+2)`
Vì `n+2\vdots n+2`
`⇒ 7\vdots n+2`
`⇒ n+2\in Ư(7)=\{±1;±7\}`
`· n+2=-1⇒n=-3`
`· n+2=1⇒n=-1`
`· n+2=-7⇒n=-9`
`· n+2=5⇒n=3`
Vậy `n\in \{±3;-1;-9\}`