Решение задания 4
$$\frac{2}{x-4} - \frac{x+8}{x^2 - 16} - \frac{1}{x} = \frac{2}{x-4} - \frac{x+8}{(x-4)(x+4)} - \frac{1}{x} = \frac{2x(x+4) - x(x+8) - (x-4)(x+4)}{x(x-4)(x+4)} = \frac{2x^2 + 8x - x^2 - 8x - (x^2 - 16)}{x(x-4)(x+4)} = \frac{2x^2 + 8x - x^2 - 8x - x^2 + 16}{x(x-4)(x+4)} = \frac{16}{x(x-4)(x+4)} = \frac{16}{x(x^2 - 16)}$$
Ответ: $$rac{16}{x(x^2 - 16)}$$