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Решите задачу:

$\frac{3}{x + 2} + \frac{2}{x - 2} = \frac{13}{ {x}^{2} - 4} ; \: x \neq -2, \: x \neq 2\\ \\ \frac{3}{x + 2} + \frac{2}{x - 2} = \frac{13}{ {x}^{2} - 4} = 0 \\ \\ \frac{3}{x + 2} + \frac{2}{x - 2} - \frac{13}{(x - 2)(x + 2)} = 0 \\ \\ \frac{3(x - 2) + 2(x + 2) - 13}{(x - 2)(x + 2)} = 0 \\ \\ \frac{3x - 6 + 2x + 4 - 13}{(x - 2)(x + 2)} = 0 \\ \\ \frac{5x + 15}{(x - 2)(x + 2)} = 0 \\ \\ 5x - 15 = 0 \\ \\ 5x = 15 \\ \\ x = 15 \div 5 \\ \\ x = 3; \: x \neq -2, \: x \neq 2$