Решите уравнение 4/(x^2-6x+9)-6/(x^2-9)=1/(x+3).

\[\frac{4}{x^{2} - 6x + 9} - \frac{6}{x^{2} - 9} = \frac{1}{x + 3}\]

\[\frac{4}{(x - 3)^{2}} - \frac{6}{(x - 3)(x + 3)} - \frac{1}{x + 3} = 0\]

\[ОДЗ:\ \ x \neq 3;\ \ x \neq - 3.\]

\[4 \cdot (x + 3) - 6 \cdot (x - 3) - (x - 3)^{2} = 0\]

\[4x + 12 - 6x + 18 - x^{2} + 6x - 9 = 0\]

\[- x^{2} + 4x + 21 = 0\]

\[x^{2} - 4x - 21 = 0\]

\[x_{1} + x_{2} = 4;\ \ \ x_{1} \cdot x_{2} = - 21\]

\[x_{1} = 7;\ \ \ x_{2} = - 3\ (не\ подходит).\]

\[Ответ:x = 7.\]