8) \(\frac{1}{x-4} - \frac{3}{x^2+4x+16} = \frac{9x+12}{x^3-64}\);

8) \(\frac{1}{x-4} - \frac{3}{x^2+4x+16} = \frac{9x+12}{x^3-64}\);

ОДЗ: \(x
eq 4\)

\(\frac{1}{x-4} - \frac{3}{x^2+4x+16} = \frac{9x+12}{(x-4)(x^2+4x+16)}\)

\(\frac{x^2+4x+16 - 3(x-4)}{(x-4)(x^2+4x+16)} = \frac{9x+12}{(x-4)(x^2+4x+16)}\)

\(x^2 + 4x + 16 - 3x + 12 = 9x + 12\)

\(x^2 + x + 28 = 9x + 12\)

\(x^2 - 8x + 16 = 0\)

\((x-4)^2 = 0\)

\(x = 4\)

Так как \(x
eq 4\), то решения нет.

Ответ: нет решений