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

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

\[ОДЗ:\ \ x \neq \frac{1}{2}\]

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

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

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

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

\[D = b^{2} - 4ac =\]

\[= 4 - 4 \cdot 1 \cdot ( - 3) =\]

\[= 4 + 12 = 16\]

\[x_{1} = \frac{- 2 + 4}{2} = \frac{2}{2} = 1\]

\[x_{2} = \frac{- 2 - 4}{2} = - \frac{6}{2} = - 3\]

\[Ответ:x = 1\ \ и\ \ x = - 3.\]