Решите уравнение (2x-10)/(x^2-3x)-(x+4)/(x^2+3x)=2/(9-x^2 ).

\[\frac{2x - 10}{x^{2} - 3x} - \frac{x + 4}{x^{2} + 3x} = \frac{2}{9 - x^{2}}\]

\[\frac{2x - 10^{\backslash x + 3}}{x(x - 3)} - \frac{x + 4^{\backslash x - 3}}{x(x + 3)} + \frac{2^{\backslash x}}{(x - 3)(x + 3)} = 0\]

\[\frac{2x^{2} - 10x + 6x - 30 - x^{2} - 4x + 3x + 12 + 2x}{x(x - 3)(x + 3)} = 0\]

\[\frac{x^{2} - 3x - 18}{x(x - 3)(x + 3)} = 0;\ \ x \neq 0;\ \ x \neq \pm 3\]

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

\[x_{1} + x_{2} = 3;\ \ \ x_{1} \cdot x_{2} = - 18\]

\[x_{1} = 6;\ \ x_{2} = - 3\ (не\ подходит).\]

\[Ответ:\ x = 6.\]