Помогите.Алгебра 8 класс

Решите задачу:

$( \frac{x}{ x^{2} -8x+16} - \frac{x+6}{ x^{2} -16} ): \frac{x+12}{ x^{2} -16} = \frac{x}{x-4} \\ \\ 1) \frac{x}{ x^{2} -8x+16} - \frac{x+6}{ x^{2} -16} = \frac{x}{(x-4)(x-4)} - \frac{x+6}{(x-4)(x+4)} = \frac{x(x+4)-(x+6)(x-4)}{(x-4)(x-4)(x+4)} = \\ \\ = \frac{ x^{2} +4x-( x^{2} -4x+6x-24)}{(x-4)(x-4)(x+4)} = \frac{ x^{2} +4x-( x^{2} +2x-24)}{(x-4)(x-4)(x+4)} = \frac{ x^{2} +4x- x^{2} -2x+24}{(x-4)(x-4)(x+4)} = \\ \\ = \frac{2x+24}{(x-4)(x-4)(x+4)} = \frac{2(x+12)}{(x-4)(x-4)(x+4)} \\ \\$

$2) \frac{2(x+12)}{(x-4)(x-4)(x+4)} : \frac{x+12}{ x^{2} -16} = \frac{2(x+12)(x-4)(x+4)}{(x-4)(x-4)(x+4)(x+12)} = \frac{2}{x-4}$