1/ \(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17=0\)
\(\Leftrightarrow\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17=0\)
\(\Leftrightarrow\left(20x^2+18x-14\right)\left(20x^2+18x+4\right)+17=0\)
Đặt t= \(20x^2+18x+4\left(t\ge0\right)\) ta có:
(t-18).t +17=0
\(\Leftrightarrow t^2-18t+17=0\)
\(\Leftrightarrow\left(t-17\right)\left(t-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=17\left(tm\right)\\t=1\left(tm\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}20x^2+18x+4=17\\20x^2+18x+4=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}20x^2+18x-13=0\\20x^2+18+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(20x+9-\sqrt{341}\right)\left(20x+9+\sqrt{341}\right)=0\\\left(20x+9-\sqrt{21}\right)\left(20x+9+\sqrt{21}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{341}}{20}\\x=\dfrac{-9-\sqrt{341}}{20}\\x=\dfrac{-9+\sqrt{21}}{20}\\x=\dfrac{-9-\sqrt{21}}{20}\end{matrix}\right.\)
Vậy.....