Đáp án:
`n in {-3, -2 , 0}`
Giải thích các bước giải:
Đặt `A = n^4 - 2n^3 + 2n^2 - 2n + 1`
`= (n^4 - n^3)-(n^3 - n^2)+ (n^2 - n)-(n-1)`
`= n^3(n-1)-n^2(n-1)+n(n-1)-(n-1)`
`= (n-1)(n^3-n^2 +n -1)`
`= (n-1)^2 (n^2+1)`
Đặt `B = n^4 -1`
`= (n-1)(n+1)(n^2+1)`
`A vdots B => n ne± 1`
`=> A vdots B <=> n-1 vdots n+1`
`<=> (n+1)-2 vdots (n+1)`
`<=> 2 vdots n+1<=>` \(\left[ \begin{array}{l}n+1=-2\\n+1=-1\\n+1=1\\n+1=2\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}n=-3\\n = -2\\n=0\\n=1 (KTM)\end{array} \right.\)
Vậy `n in {-3,-2,0}` thì `n^4 - 2n^3 + 2n^2 - 2n + 1 vdots n^4 -1`
