\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)-\left(2x+1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left(24x^2+16x-9x-6\right)-\left(4x^2+16x+7x+28\right)-\left(10x^2-2x+5x-1\right)=0\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28-10x^2+2x-5x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Leftrightarrow10x^2+11x-30x-33=0\)
\(\Leftrightarrow\left(10x^2+11x\right)-\left(30x+33\right)=0\)
\(\Leftrightarrow x\left(10x+11\right)-3\left(10x+11\right)=0\)
\(\Leftrightarrow\left(10x+11\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}10x+11=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}10x=-11\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11}{10}\\x=3\end{matrix}\right.\)
Vậy \(x=\dfrac{-11}{10}\) hoặc \(x=3\)
