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

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

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

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

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

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

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

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

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

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

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

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

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

\[3 \cdot \left( x + \frac{2}{3} \right)^{2} + \frac{3}{4} > 0\ \ \ и\ \ \ \]

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

\[решения.\]

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

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

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

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

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

\[Ответ:( - \infty; - 2) \cup (0;1\rbrack\]