Решите уравнение: 3/(x^2+4x)-15/(x^2-4x)=4/x.

\[\frac{3}{x^{2} + 4x} - \frac{15}{x^{2} - 4x} = \frac{4}{x}\]

\[\frac{3}{x(x + 4)} - \frac{15}{x(x - 4)} = \frac{4}{x}\]

\[3 \cdot (x - 4) - 15 \cdot (x + 4) = 4 \cdot (x^{2} - 16)\]

\[ОДЗ:x \neq 0;x \neq \pm 4.\]

\[3x - 12 - 15x - 60 = 4x^{2} - 64\]

\[4x^{2} + 12x + 8 = 0\ \ \ |\ :4\]

\[x^{2} + 3x + 2 = 0\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = 2\]

\[x_{1} = - 2;\ \ x_{2} = - 1.\]

\[Ответ:x = - 2;x = - 1.\]