Решите неравенство: (x^2-16)/(x+1)<=0.

\[\frac{x^{2} - 16}{x + 1} \leq 0\]

\[\frac{(x - 4)(x + 4)}{x + 1} \leq 0\]

\[Ответ:( - \infty; - 4\rbrack \cup ( - 1;4\rbrack.\]