Решите уравнение: 1/(x-3)-2/(x^2+3x+9)=(6+7x)/(x^3-27).

\[\frac{1}{x - 3} - \frac{2}{x^{2} + 3x + 9} = \frac{6 + 7x}{x³ - 27}\]

\[x^{2} + 3x + 9 - 2 \cdot (x - 3) =\]

\[= 6 + 7x\]

\[x^{2} + 3x + 9 - 2x + 6 - 6 - 7x =\]

\[= 0\]

\[x^{2} - 6x + 9 = 0\]

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

\[x - 3 = 0\]

\[x = 3\ (не\ подходит).\]

\[Ответ:нет\ решения.\]