a, \(\left(3x-2\right).\left(2x-3\right)-x.\left(6x-4\right)=11\)
\(\Leftrightarrow6x^2-9x-4x+6-6x^2+4x=11\)
\(\Leftrightarrow-9x^2+6=11\)
\(\Leftrightarrow-9x=5\)
\(\Leftrightarrow x=\dfrac{-5}{9}\)
b, \(x\left(x-3\right)-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(2x-6\right)=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy \(x=3\text{ hoặc }x=2\)
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