Решите неравенство: x^2/(x-3)>(-9)/(3-x).

\[\frac{x^{2}}{x - 3} > \frac{- 9}{3 - x}\]

\[\frac{x^{2}}{x - 3} > \frac{9}{x - 3}\]

\[\frac{x^{2} - 9}{x - 3} > 0\]

\[\frac{(x - 3)(x + 3)}{x - 3} > 0\]

\[Ответ:( - 3;3) \cup (3; + \infty).\]