Решите уравнение: 27/(x^2+3x)-2/x=3/(x^2-3x).

\[\frac{27}{x^{2} + 3x} - \frac{2}{x} = \frac{3}{x^{2} - 3x}\]

\[\frac{27^{\backslash x - 3}}{x(x + 3)} - \frac{2^{\backslash x^{2} - 9}}{x} = \frac{3^{\backslash x + 3}}{x(x - 3)}\]

\[ОДЗ:\ \ x \neq 0;\ \ x \neq \pm 3.\]

\[27x - 81 - 2x^{2} + 18 = 3x + 9\]

\[2x^{2} - 24x + 72 = 0\ \ \ \ |\ :2\]

\[x^{2} - 12x + 36 = 0\]

\[(x - 6)^{2} = 0\]

\[x - 6 = 0\]

\[x = 6.\]

\[Ответ:x = 6.\]