~ Bạn tham khảo ~
`D= [x^2-1]/[x+1] = [x^2+x-x-1]/[x+1] = [x(x+1)-(x+1)]/[x+1]=[(x+1)(x-1)]/[x+1]=x-1`
Để `D` nguyên
`=> x-1` nguyên
`=> x-1 \in {-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;....}`
`=> x \in {-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7;.....}`
Vậy `D` nguyên với mọi `x`