`\text{Bài 5:}`
`a)` Ta có: `(2n+1)/(n-3)=(2(n-3)+7)/(n-3)=2+7/(n-3)`
Do `2n+1 ⋮ n-3` nên `2+7/(n-3)⋮n-3`
`⇒7⋮n-3`
`⇒n-3∈Ư(7)={±1;±7}`
Mà `n∈N` nên `n-3∈{1;7}`
`⇒n∈{4;10}`
Vậy `n∈{4;10}`
`b)` Ta có:
`(n^2+3)/(n+1)=(n(n+1)-(n-3))/(n+1)=n-(n-3)/(n+1)=n-(n+1-4)/(n+1)=n-1-4/(n+1)`
Do `n^2+3⋮ n+1` nên `n-1-4/(n+1)⋮n+1`
`⇒4⋮n+1`
`⇒n+1∈{±1;±2;±4}`
Mà `n∈N` nên `n+1∈{1;2;4}`
`⇒ n∈{0;1;3}`
Vậy `n∈{0;1;3}`.
`\text{Bài 6:}`
`a)` `1+2+3+4+...+n=231`
`⇒(n(n+1))/2=231`
`⇒n(n+1)=462`
`⇒n(n+1)=2.3.7.11=21.22`
`⇒n=21`