Помогите пожалуйста, заранее спасибо

Решите задачу:

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

$\frac{x^2+3x}{4ax-16a} : \frac{x+3}{3x-12} = \frac{x(x+3) }{4a(x-4)} * \frac{3(x-4)}{x+3} = \frac{3x}{4a} \\ \\ \frac{9x^2-1}{6x^2-2x} = \frac{(3x)^2-1^2}{2x(3x-1)} = \frac{(3x-1)(3x+1)}{2x(3x-1)} = \frac{3x+1}{2x} \\ \\ \frac{x^2-4}{x^2-4x+4} = \frac{x^2-2^2}{x^2-2*2*x +2^2} = \frac{(x-2)(x+2)}{(x-2)^2} = \frac{x+2}{x-2}$

$\frac{x^2+14x+49}{x^2+8x+7} = \frac{x^2+2*7*x +7^2}{x^2+7x+x+7} = \frac{(x+7)^2}{x(x+7) +1(x+7)} = \frac{(x+7)(x+7)}{(x+1)(x+7)} = \frac{x+7}{x+1} \\ \\ ( \frac{1}{x+3} - \frac{2}{x^2-9} : \frac{x}{x-3} ) : (1 - \frac{2}{x} )=( \frac{1}{x+3} - \frac{2}{(x-3)(x+3)} * \frac{x-3}{x} ) : ( \frac{1*x-2}{x} )= \\ =( \frac{1}{x+3} - \frac{2}{x(x+3)} ) * \frac{x}{x-2} = \frac{x-2}{x(x+3)} * \frac{x}{x-2} = \frac{1}{x+3}$