Решите уравнение: 1/(x-4)-3/(x^2+4x+16)=(9x+12)/(x^3-64).

\[\frac{1}{x - 4} - \frac{3}{x^{2} + 4x + 16} = \frac{9x + 12}{x^{3} - 64}\]

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

\[= \frac{9x + 12}{(x - 4)\left( x^{2} + 4x + 16 \right)};\ \ \ \ \ \]

\[x \neq 4\]

\[x^{2} - 5x + 4 - 3x + 12 = 0\]

\[x^{2} - 8x + 16 = 0\]

\[(x - 4)^{2} = 0\]

\[x - 4 = 0\ \ \]

\[x = 4\ \ (не\ подходит).\]

\[Ответ:нет\ решения.\]