$\\$
Bạn tham khảo bài.
`P=(6n - 7)/(2n+1)(n \ne (-1)/2)`
Để `P` nguyên
`-> 6n-7 \vdots 2n+1`
`->6n+3 - 10 \vdots 2n+1`
`-> 3 (2n+1) - 10 \vdots 2n+1`
Vì `2n+1 \vdots 2n+1 ->3(2n+1) \vdots 2n+1`
`-> -10 \vdots 2n+1`
`->2n+1 ∈ Ư (-10)={1;-1;2;-2;5;-5;10;-10}`
`->2n ∈ {0;-2;1;-3;4;-6;9;-11}`
`-> n ∈ {0; -1;1/2; (-3)/2; 2;-3;9/2;(-11)/2}`
Vì `n∈ZZ`
`->n ∈ {0;-1;2;-3}`
Vậy `n∈{0;-1;2;-3}` để `P` nguyên
