(2a/(2a+b) - 4a^2/(4^2+4ab+b^2)) / (2a/(4a^2-b^2) + 1/(b-2a)

Решите задачу:

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