Решите систему неравенств: (x+3)/(x+1)>=0; x^2+3x-4<=0.

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

\[x^{2} + 3x - 4 = x^{2} + 4x - x - 4 =\]

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

\[= (x + 4)(x - 1)\]

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

\[Ответ:x \in \lbrack - 4; - 3\rbrack \cup ( - 1;1\rbrack.\]