Решите с объяснением,пожалуйста.

Решите задачу:

${} \ \ \ \displaystyle \left(3x-\frac{3x}{x-4} \right) : \left(x-\frac{6x-25}{x-4} \right) = \\\\\\ =\frac{3x(x-4)-3x}{x-4} :\frac{x(x-4)-(6x-25)}{x-4} = \\\\\\ =\frac{3x^2-12x-3x}{x-4} :\frac{x^2-4x-6x+25}{x-4} = \\\\\\ = \frac{3x^2-15x}{x-4} :\frac{x^2-10x+25}{x-4}= \frac{3x(x-5)}{x-4} :\frac{(x-5)^2}{x-4}= \\\\\\ =\frac{3x(x-5)}{x-4} \cdot \frac{x-4}{(x-5)^2}= \frac{3x}{x-5}$