Решите уравнение: 1/(x-4)-3/(x^2+4x)=24/(x^3-16x).

\[\frac{1}{x - 4} - \frac{3}{x^{2} + 4x} = \frac{24}{x³ - 16x}\]

\[\frac{1}{x - 4} - \frac{3}{x(x + 4)} = \frac{24}{x(x^{2} - 16)}\ \]

\[x(x + 4) - 3 \cdot (x - 4) = 24\]

\[x^{2} + 4x - 3x + 12 - 24 = 0\]

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

\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 12\]

\[x_{1} = 3,\ \ \]

\[x_{2} = - 4\ (не\ подходит)\]

\[Ответ:x = 3.\]