Найдите корни уравнения 3/(x^2+4x)-15/(x^2-4x)=4/x.

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

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

\[\frac{3x - 12 - 15x - 60 - 4x^{2} + 64}{x\left( x^{2} - 16 \right)} = 0\]

\[ОДЗ:\ \ x \neq 0;\ \ \ x \neq \pm 4\]

\[- 4x^{2} - 12x - 8 = 0\ \ \ \ \ |\ :( - 4)\]

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

\[D = 9 - 8 = 1\]

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

\[Ответ:x = - 1;x = - 2.\]