Đặt `t=x^2+x+1`
`PT⇒t(t+11)=12`
`⇔t^2+11t-12=0`
`⇔t^2-t+12t-12=0`
`⇔t(t-1)+12(t-1)`
`⇔(t+12)(t-1)=0`
`⇔`\(\left[ \begin{array}{l}t+12=0\\t-1=0\end{array} \right.\)
TH1: `t+12=0`
`⇔x^2+x+13=0`
`⇔x^2+x+1/4=-51/4`
`⇔(x+1/2)^2=-51/4` (vô lý)
TH2:
`t-1=0`
`⇔x^2+x+1-1=0`
`⇔x^2+x=0`
`⇔x(x+1)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)