Помогите пожалуйста, 2С.55(а).

Решите задачу:

$(\frac{3}{x-4} + \frac{4x-6}{x^2-3x-4} + \frac{2x}{x+1}) *\frac{x}{2x-3} = \\ \\ = (\frac{3}{x-4} + \frac{4x-6}{x^2+x-4x-4} + \frac{2x}{x+1}) *\frac{x}{2x-3} = \\ \\ =(\frac{3}{x-4} + \frac{4x-6}{x(x+1)-4(x+1)} + \frac{2x}{x+1}) *\frac{x}{2x-3} = \\ \\ =(\frac{3(x+1)}{(x-4)(x+1)} + \frac{4x-6}{(x+1)(x-4)} + \frac{2x(x-4)}{(x+1)(x-4)}) *\frac{x}{2x-3} = \\ \\ = \frac{3x+3+4x-6+2x^2-8x}{(x-4)(x+1)} *\frac{x}{2x-3} = \\ \\ = \frac{2x^2-x-3}{(x-4)(x+1)} *\frac{x}{2x-3}=$

$= \frac{2(x^2-0.5x-1.5)}{(x-4)(x+1)} *\frac{x}{2(x-1.5)}= \frac{2(x^2+x-1.5x-1.5)}{(x-4)(x+1)} *\frac{x}{2(x-1.5)}= \\ \\ =\frac{2(x(x+1)-1.5(x+1))}{(x-4)(x+1)} *\frac{x}{2(x-1.5)}= \frac{2x(x-1.5)(x+1)}{2(x-1.5)(x-4)(x+1) } = \\ \\ = \frac{x}{x-4}$