Найдите корни уравнения: 1/(x-3)^2 +9/(x+3)^2 -6/(x^2-9)=0.

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

\[= 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[ОДЗ:\ \ x \neq \pm 3\]

\[4x² - 48x + 144 = 0\ \ \ \ |\ :4\]

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

\[D = b^{2} - 4ac =\]

\[= 144 - 4 \cdot 1 \cdot 36 =\]

\[= 144 - 144 = 0\]

\[x = \frac{12}{2} = 6\]

\[Ответ:x = 6.\]