\frac{2x}{x^{2}-64} -\frac{1}{x-8} при x=-4 \frac{2x}{x^{2}-9} -\frac{1}{x+3} при x=-5

Решите задачу:

$1) \frac{2x}{x^2-64} - \frac{1}{x-8} = \frac{2x}{(x-8)(x+8)} - \frac{1}{x-8} = \frac{2x-(x+8)*1}{(x-8)(x+8)} = \frac{2x-x-8}{( x-8)(x+8)} = \\ \\ = \frac{x-8}{x-8)(x+8)} = \frac{1}{x+8} = \frac{1}{-4+8}= \frac{1}{4} =0,25 \\ \\ \\ \\ 2) \frac{2x}{x^2-9}- \frac{1}{x+3}= \frac{2x}{(x-3)(x+3)} - \frac{1}{x+3}= \frac{2x-(x-3)*1}{(x-3)(x+3)} = \frac{2x-x+3}{(x-3)(x+3)} = \\ \\ = \frac{x+3}{(x-3)(x+3)} = \frac{1}{x-3} = \frac{1}{-5-3} =- \frac{1}{8} =-0,125$