Đáp án:
`x in {0,1,-2,3}`
Giải thích các bước giải:
`7/(x^2-x+1) in Z`
`=>7 vdots x^2-x+1`
`=>x^2-x+1 in Ư(7)={+-1,+-7}`
Mà `x^2-x+1`
`=x^2-2.x.1/2+1/4+3/4`
`=(x-1/2)^2+3/4>=3/4>0`
`=>x^2-x+1 in {1,7}`
`+)x^2-x+1=1`
`=>x^2-x=0`
`=>` \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
`+)x^2-x+1=7`
`=>x^2-x-6=0`
`=>x^2+2x-3x-6=0`
`=>x(x+2)-3(x-2)=0`
`=>(x+2)(x-3)=0`
`=>` \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
Vậy với `x in {0,1,-2,3}` thì `7/(x^2-x+1) in Z`
