Решите неравенство: (x^2+3x+2)/(x^2+x+1)-(6x^2+6x+6)/(x^2+3x+2)+1<=0.

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

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

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

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

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

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

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

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

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

\[\frac{4x^{2} + 6x + 5}{x^{2} + x + 1} \leq 0\]

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

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

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

\[решения.\]

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

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

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

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