(x²-x+1)(x²-x+2)=2
Đặt x² - x + 1 = a
⇒ a (a+1)=2
⇔a²+a-2=0
⇔a²-a+2a-2=0
⇔a(a-1)+2(x-1)=0
⇔(a-1)(a+2)=0
⇔\(\left[ \begin{array}{l}x^2-x+1-1=0\\x^2-x+1+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x^2-x=0\\x^2-x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x(x-1)=0\\x^2-x+\frac{1}{4}+\frac{11}{4}=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0;x-1=0\\(x-\frac{1}{2})^2=-\frac{11}{4}(loại)\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Vậy x∈{ 0 ; 1 }
