Решите неравенство: (x^2+x)/(x^2-x+1)-(6x^2-6x+6)/(x^2+x)+1<=0.

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

\[t - \frac{6}{t} + 1 \leq 0\]

\[\frac{t^{2} + t - 6}{t} \leq 0\]

\[\frac{(t - 2)(t + 3)}{t} \leq 0\]

\[t \leq - 3\ \ и\ \ 0 < t \leq 2 \Longrightarrow\]

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

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

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

\[\frac{x^{2} + x + 3 \cdot (x^{2} - x + 1)}{x^{2} - x + 1} \leq 0\]

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

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

\[\frac{4 \cdot \left( x - \frac{1}{4} \right)^{2} + \frac{11}{4}}{\left( x - \frac{1}{2} \right)^{2} + \frac{3}{4}} \leq 0\]

\[4 \cdot \left( x - \frac{1}{4} \right)^{2} + \frac{11}{4} > 0\ \ \ \]

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

\[решений.\]

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

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

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

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

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

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