\(\left(2x-2\right).\left(3x-9\right)< 0\Leftrightarrow2\left(x-1\right).3\left(x-3\right)< 0\)
\(\Leftrightarrow6\left(x-1\right)\left(x-3\right)< 0\Leftrightarrow\orbr{\begin{cases}x-1< 0;x-3>0\\x-1>0;x-3< 0\end{cases}}\)
Mà \(x-1>x-3\Rightarrow\hept{\begin{cases}x-1>0\\x-3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x< 3\end{cases}}\Leftrightarrow1< x< 3\Leftrightarrow x=2\)
Vậy \(x=2\)