(a-3d)//(a-2b)-(3a-b)//(a+2b)+(6ab-16b^2)//(4b^2-a^2)

Решите задачу:

$\frac{a-3b}{a-2b}-\frac{3a-b}{a+2b}+\frac{6ab-16b^2}{4b^2-a^2}=\frac{a-3b}{a-2b}-\frac{3a-b}{a+2b}-\frac{6ab-16b^2}{(a-2b)(a+2b)}=\\\ =\frac{(a-3b)(a+2b)-(3a-b)(a-2b)-(6ab-16b^2)}{(a-2b)(a+2b)}=\\\ =\frac{a^2-ab-6b^2-3a^2+7ab-2b^2-6ab+16b^2}{(a-2b)(a+2b)}=\\\ =\frac{-2a^2+8b^2}{a^2-4b^2}=-\frac{2(a^2-4b^2)}{a^2-4b^2}=-2$