Решите уравнение: 2x/(x^2-3x+1)-2/(x^3-2x^2+x)=7/(3x^2-3x).

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

\[= \frac{7}{3x² - 3x}\]

\[2x \cdot 3x - 2 \cdot 3 - 7 \cdot (x - 1) = 0\]

\[6x^{2} - 6 - 7x + 7 = 0\]

\[6x^{2} - 7x + 1 = 0\]

\[D = ( - 7)^{2} - 4 \cdot 6 \cdot 1 =\]

\[= 49 - 24 = 25;\ \ \ \ \sqrt{D} = 5.\]

\[x_{1} = \frac{7 + 5}{2 \cdot 6} = \frac{12}{12} =\]

\[= 1\ (не\ подходит);\ \ \ \ \ \ \ \]

\[x_{2} = \frac{7 - 5}{2 \cdot 6} = \frac{2}{12} = \frac{1}{6}\]

\[Ответ:\ \frac{1}{6}.\]