\(\left(6x-3\right)\left(2x-7\right)-\left(3x-4\right)\left(3x+2\right)=\left(x-3\right)\left(3x+1\right)\)
\(\Leftrightarrow\left(6x-3\right)\left(2x-7\right)-\left(3x-4\right)\left(3x+2\right)-\left(x-3\right)\left(3x+1\right)=0\)
\(\Leftrightarrow12x^2-6x-42x+21-\left(9x^2-12x+6x-8\right)-\left(3x^2-9x+x-3\right)=0\)
\(\Leftrightarrow12x^2-48x+21-\left(9x^2-6x-8\right)-\left(3x^2-8x-3\right)=0\)
\(\Leftrightarrow12x^2-48x+21-9x^2+6x+8-3x^2+8x+3=0\)
\(\Leftrightarrow-34x+32=0\)
\(\Leftrightarrow-34x=-32\)
\(\Leftrightarrow x=\dfrac{16}{17}\)
Vậy \(x=\dfrac{16}{17}\)
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