Question

(3a+3b/a^2-ab+b^2-3/a+b)÷3ab/a^3+b^3=3 доказать

Answer 1

$............= frac{3(a+b)}{a^{2}-ab+b^{2} }* frac{(a+b)(a^{2}-ab+b^{2})}{3ab}- frac{3}{a+b}* frac{(a+b)(a^{2}-ab+b^{2})}{3ab}= \ \ frac{(a+b)(a+b)}{ab} - frac{(a^{2}-ab+b^{2})}{ab}= frac{a^{2}+2ab+b^{2}-a^{2}-ab+b^{2}}{ab}= frac{3ab}{ab} =3$

3=3 - верно