`(x^2-x)/(x^2+x+1)-(x^2-x+2)/(x^2-x-2)=1`
ĐKXĐ: `xne2;xne-1`
`⇔(x^2-x+1-1)/(x^2-x+1)-(x^2-x-2+4)/(x^2-x-2)=1`
`⇔(x^2-x+1)/(x^2-x+1)-1/(x^2-x+1)-(x^2-x-2)-4/(x^2-x-2)=1`
`⇔1-1/(x^2-x+1)-1-4/(x^2-x-2)=1`
`⇔1/(x^2-x-2+3)+4/(x^2-x-2)+1=0`
Đặt `x^2-x-2=t (ĐKXĐ:xne0;xne-3)`
`1/(t+3)+4/t+1=0`
`⇔t/(t(t+3))+(4(t+3))/(t(t+3))+(t(t+3))/(t(t+3))=0`
`⇔t+4t+12+t^2+3t=0`
`⇔t^2+8t+12=0`
`⇔(t+2)(t+6)=0`
`⇔`\(\left[ \begin{array}{l}t=-2(nhận)\\t=-6(nhận)\end{array} \right.\)
Với `t=-2` ta có:
`x^2-x-2=-2`
`⇔x^2-x=0`
`⇔x(x-1)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Với `t=-6` ta có:
`x^2-x-2=-6`
`⇔x^2-x+4=0`
`⇔x^2-2x.1/2+1/4+15/4=0`
`⇔(x-1/2)^2+15/4=0`
Do `(x-1/2)^2≥0∀xne2;xne-1`
`⇔(x-1/2)^2+15/4≥15/4>0 (Loại)`
Vậy `S={0;1}`
