Помогите пожалуйста

Решите задачу:

$\frac{2x^2-8y^2}{3x-y} * \frac{6x-2y}{5x+10y}= \frac{2(x-2y)(x+2y)}{3x-y} * \frac{2(3x-y)}{5(x+2y)}= \frac{4}{5} (x-2y) \\ \frac{4}{5} (x-2y) = \frac{4}{5}(9-2*(- \frac{1}{2})=8$

$\frac{a^2+6ab+9b^2}{b^2+4}* \frac{b^4+4b^2}{a^2-9b^2} = \frac{(a+3b)^2}{b^2+4}* \frac{b^2(b^2+4)}{(a-3b)(a+3b)} = \frac{b^2(a+3b)}{a-3b} \\\frac{b^2(a+3b)}{a-3b} =\frac{(-0,5)^2(1,5+3(-0,5))}{1,5-3(-0,5)} =0$