Найдите корни уравнения: 3/(x+2)-3/(2-x)=2/(x^2-4).

\[\frac{3}{x + 2} - \frac{3}{2 - x} = \frac{2}{x² - 4}\]

\[ОДЗ:\ \ x \neq \pm 2\]

\[\frac{3}{x + 2} + \frac{3}{x - 2} = \frac{2}{x² - 4}\]

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

\[3x - 6 + 3x + 6 = 2\]

\[6x = 2\]

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

\[x = \frac{1}{3}\]

\[Ответ:\ \ x = \frac{1}{3}.\]