Решите систему неравенств: (x^2-4x+4)/(x+2)<=0; x^2+2x-15<=0.

\[\left\{ \begin{matrix} \frac{x^{2} - 4x + 4}{x + 2} \leq 0\ \ \\ x^{2} + 2x - 15 \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

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

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

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

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

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

\[Ответ:\lbrack - 5; - 2) \cup \left\{ 2 \right\}.\]