(2x²+x+1).(2x²+x-4) = -4
⇔ (2x²+x+1).(2x²+x-4) + 4 = 0
Đặt c = 2x² +x +1
c.(c -5) +4 = 0
⇔ c² -5c +4 = 0
⇔ c² -4c -c +4 = 0
⇔ c.(c -4) - (c -4) = 0
⇔ (c -4).(c -1) = 0
⇔ \(\left[ \begin{array}{l}c -4=0\\c -1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}c=4\\c=1\end{array} \right.\)
TH1: Thay c = 4 vào 2x² +x +1, ta đc:
2x² +x +1 = 4
⇔ 2x² +x +1 -4 = 0
⇔ 2x² +x -3 = 0
⇔ 2x² +3x -2x -3 = 0
⇔ x.(2x +3) - (2x +3) = 0
⇔ (2x +3).(x -1) = 0
⇔ \(\left[ \begin{array}{l}2x+3=0\\x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{-3}{2}\\x=1\end{array} \right.\)
TH2: thay a = 1 vào 2x² +x +1, ta đc:
2x² +x +1 = 1
⇔ 2x² +x +1 -1 = 0
⇔ 2x² +x = 0
⇔ x.(2x +1) = 0
⇔ \(\left[ \begin{array}{l}x=0\\2x+1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=\frac{-1}{2}\end{array} \right.\)
Vậy S= { $\frac{-1}{2}$ ; 0; 1; $\frac{-3}{2}$ }