Решите уравнение: (x^2-6x)/(3x-1)=(3x-4)/(1-3x).

\[\frac{x^{2} - 6x}{3x - 1} = \frac{3x - 4}{1 - 3x}\]

\[ОДЗ:\ \ x \neq \frac{1}{3}\ \]

\[\frac{x² - 6x}{3x - 1} = - \frac{3x - 4}{3x - 1}\]

\[\frac{x² - 6x + 3x - 4}{3x - 1} = 0\]

\[\frac{x^{2} - 3x - 4}{3x - 1} = 0\]

\[x^{2} - 3x - 4 = 0\]

\[x_{1} + x_{2} = 3\]

\[x_{1} \cdot x_{2} = - 4 \Longrightarrow x_{1} = 4;\ \ x_{2} = - 1\]

\[Ответ:\ \ x = 4\ \ \ \ и\ \ \ x = - 1.\]