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

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

\[ОДЗ:\ \ x \neq \frac{1}{2}\]

\[\frac{2x² + x - 1 - 2 \cdot (2x - 1)}{2x - 1} = 0\]

\[\frac{2x² + x - 1 - 4x + 2}{2x - 1} = 0\]

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

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

\[D = b^{2} - 4ac = 9 - 4 \cdot 2 =\]

\[= 9 - 8 = 1\]

\[x_{1} = \frac{3 + 1}{4} = \frac{4}{4} = 1\]

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

\[= \frac{1}{2}\ \ \ (не\ подходит\ по\ ОДЗ)\]

\[Ответ:\ \ x = \ 1.\]