A cura di: Nicola Vitale

$(frac{a^2}{4a^2+4ab+b^2} – frac{a-b}{6a+3b}):(frac{a^3-b^3}{12a+6b])$


Soluzione

$frac{a^2}{(2a+b)^2} – frac{a-b}{3  (2a+b)}) : frac{a^3-b^3}{6  (2a+b)}$
$frac{3a^2-(a-b)(2a+b)}{3  (2a+b)^2}:frac{a^3-b^3}{6  (2a+b)}$
$frac{3a^2-(2a^2+ab-2ab-b^2)}{3  (2a+b)^2} cdot frac{6  (2a+b)}{a^3-b^3}$
$frac{3a^2-2a^2-ab+2ab+b^2}{3  (2a+b)^2} cdot frac{6  (2a+b)}{a^3-b^3}$
$frac{1}{(2a+b)} cdot frac{2}{(a-b)}$
$frac{2}{(2a+b)(a-b)}$