Помогите пожалуйста.

Решите задачу:

$1) = \frac{3n*(m-2)}{2m^3} * \frac{4m^4}{m^2n-4n} = 3n*(m-2)*\frac{2m}{n*(m^2-4}) = 3(m-2)*\frac{2m}{(m-2)*(m+2)} = 3*\frac{2m}{m+2} = \frac{6m}{m+2} \\\\2) = (2a-b)^2 * \frac{2}{(2a-b)*(2a+b)} = (2a-b) * \frac{2}{2a+b} = \frac{4a-2b}{2a+b} \\\\3) = \frac{(x-3)*(x+3)}{(x+2)^2} * \frac{x+2}{x-3} = \frac{x+3}{x+2} \\\\4) = \frac{a}{(a+2)*(a^2-2a+4)} * \frac{(a+2)^2}{a^2} = \frac{a}{a^2-2a+4} * (a+2) = \frac{a^2+2a}{a^2-2a+4}$