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

\[\left( \frac{x^{2} + 3x}{x + 2} \right)^{2} - \frac{x^{2} + 3x}{x + 2} - 6 \leq 0\]

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

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

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

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

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

\[\left\{ \begin{matrix} \frac{x^{2} + 3x + 2x + 4}{x + 2} \geq 0 \\ \frac{x^{2} + 3x - 3x - 6}{x + 2} \leq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \frac{x^{2} + 5x + 4}{x + 2} \geq 0 \\ \frac{x^{2} - 6}{x + 2} \leq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{(x + 1)(x + 4)}{x + 2} \geq 0\ \ \ \ \ \ \\ \frac{\left( x - \sqrt{6} \right)\left( x + \sqrt{6} \right)}{x + 2} \leq 0 \\ \end{matrix} \right.\ \]

\[Ответ:\left\lbrack - 4; - \sqrt{6} \right\rbrack \cup \left\lbrack - 1;\sqrt{6} \right\rbrack\text{.\ }\]