Ta có: \(\left(2-5x\right)\left(6x^2+7x-8\right)=\left(5x-2\right)\left(11+2x\right)\)
\(\Rightarrow\left(2-5x\right)\left(6x^2+7x-8\right)-\left(5x-2\right)\left(11+2x\right)=0\)
\(\Rightarrow\left(2-5x\right)\left(6x^2+7x-8\right)+\left(2-5x\right)\left(11+2x\right)=0\)
\(\Rightarrow\left(2-5x\right)\left(6x^2+7x-8+11+2x\right)=0\)
\(\Rightarrow\left(2-5x\right)\left(6x^2+9x+3\right)=0\)
\(\Rightarrow\left(2-5x\right)\cdot3\cdot\left(2x^2+3x+1\right)=0\)
\(\Rightarrow\left(2-5x\right)\cdot3\cdot\left(2x^2+2x+x+1\right)=0\)
\(\Rightarrow\left(2-5x\right)\cdot3\cdot\left[2x\left(x+1\right)+\left(x+1\right)\right]=0\)
\(\Rightarrow\left(2-5x\right)\cdot3\cdot\left(x+1\right)\left(2x+1\right)=0\)
Vì \(3\ne0\) nên
\(\Rightarrow\left[{}\begin{matrix}2-5x=0\\x+1=0\\2x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}5x=2\\x=-1\\2x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{5}\\x=-1\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{2}{5};-1;\frac{-1}{2}\right\}\)
