Решите систему неравенств: (x^2-4x+4)/(x+1)<=0; x^2+4x-12<=0.

\[\left\{ \begin{matrix} \frac{x^{2} - 4x + 4}{x + 1} \leq 0\ \ \\ x^{2} + 4x - 12 \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

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

\[x^{2} + 6x - 2x - 12 =\]

\[= x(x + 6) - 2(x + 6) =\]

\[= (x + 6)(x - 2)\]

\[Ответ:\lbrack - 6;\ - 1) \cup \left\{ 2 \right\}.\]