Решите неравенство: (2x^2)/(x-3)>(-18)/(3-x).

\[\frac{2x^{2}}{x - 3} > \frac{- 18}{3 - x}\]

\[\frac{2x^{2}}{x - 3} > \frac{18}{x - 3}\]

\[\frac{2x^{2} - 18}{x - 3} > 0\]

\[\frac{2 \cdot \left( x^{2} - 9 \right)}{x - 3} > 0\]

\[\frac{2 \cdot (x - 3)(x + 3)}{x - 3} > 0\]

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