Q(x)=(4x−2)(10x+4)(5x+7)(2x+1)+17=0(4x−2)(10x+4)(5x+7)(2x+1)+17=0
⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0
⇔(20x2+18x−14)(20x2+18x+4)+17=0⇔(20x2+18x−14)(20x2+18x+4)+17=0
Đặt t= 20x2+18x+4(t≥0)20x2+18x+4(t≥0) ta có:
(t-18).t +17=0
⇔t2−18t+17=0⇔t2−18t+17=0
⇔(t−17)(t−1)=0
⇔\(\left[ \begin{array}{l}t=17\\t=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}20x^{2}+18x+4=17\\20x^{2}+18x+4=1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}20x^{2}+18x-13=0\\20x^{2}+18x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}(20x+9-√341)(20x+9+√341)=0\\(20x+9−√21)(20x+9+√21)=0\end{array} \right.\)
