Решите неравенство: (2x^2-3x+1)/(x^2-x+1)-(2x^2-2x+2)/(2x^2-3x+1)+1<=0.

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

\[t - \frac{2}{t} + 1 \leq 0\]

\[\frac{t^{2} + t - 2}{t} \leq 0\]

\[\frac{(t - 1)(t + 2)}{t} \leq 0\]

\[t \leq - 2\ \ и\ \ 0 < t \leq 1 \Longrightarrow\]

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

\[и\ \ 0 < \frac{2x^{2} - 3x + 1}{x^{2} - x + 1} \leq 1\]

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

\[\frac{4x^{2} - 5x + 3}{x^{2} - x + 1} \leq 0\]

\[\frac{4 \cdot \left( x - \frac{5}{8} \right)^{2} + \frac{23}{16}}{\left( x - \frac{1}{2} \right)^{2} + \frac{3}{4}} \leq 0\]

\[4 \cdot \left( x - \frac{5}{8} \right)^{2} + \frac{23}{16} > 0\ \ \]

\[и\ \ \left( x - \frac{1}{2} \right)^{2} + \frac{3}{4} > 0 \Longrightarrow нет\ \]

\[решения.\]

\[\left\{ \begin{matrix} \frac{2x^{2} - 3x + 1}{x^{2} - x + 1} > 0 \\ \frac{2x^{2} - 3x + 1}{x^{2} - x + 1} \leq 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} (x - 1)(2x - 1) > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{2x^{2} - 3x + 1 - x^{2} + x - 1}{x^{2} - x + 1} \leq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2 \cdot (x - 1)\left( x - \frac{1}{2} \right) > 0 \\ \frac{x^{2} - 2x}{\left( x - \frac{1}{2} \right)^{2} + \frac{3}{4}} \leq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2 \cdot (x - 1)(x - 0,5) > 0 \\ x(x - 2) \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\lbrack 0;0,5) \cup (1;2\rbrack.\]