Решите систему неравенств: (x^2-6x+9)/(x+2)<=0; x^2-2x-15<=0.

\[\left\{ \begin{matrix} \frac{x^{2} - 6x + 9}{x + 2} \leq 0\ \ \\ x^{2} - 2x - 15 \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{(x - 3)^{2}}{x + 2} \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 5)(x + 3) \leq 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 2x - 15 =\]

\[= x^{2} + 3x - 5x - 15 =\]

\[= x(x + 3) - 5(x + 3) =\]

\[= (x + 3)(x - 5)\]

\[Ответ:\lbrack - 3;\ - 2) \cup \left\{ 3 \right\}.\]