Решите систему неравенств: (x^2-6x+9)/(x+1)<=0; x^2+x-20<=0.

\[\left\{ \begin{matrix} \frac{x^{2} - 6x + 9}{x + 1} \leq 0 \\ x^{2} + x - 20 \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{(x - 3)^{2}}{x + 1} \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 4)(x + 5) \leq 0 \\ \end{matrix} \right.\ \]

\[x^{2} + x - 20 =\]

\[= x^{2} + 5x - 4x - 20 =\]

\[= x(x + 5) - 4(x + 5) =\]

\[= (x + 5)(x - 4)\]

\[Ответ:\lbrack - 5;\ - 1) \cup \left\{ 3 \right\}.\]