因为a+b+c=0,则b=-a-c,bc=-ac-c2
所以2a2+bc=2a2-ac-c2=(2a+c)(a-c)=(a-b)(a-c)
故:[a2/(2a2+bc)]+[b2/(2b2+ac)]+[C2/(2c2+ab)]
=[a2/(a-b)(a-c)]+[b2/(b-c)(b-a)]+[C2/(c-a)(c-b)]
=(a-b)*[a2/(a-c)-b2/(b-c)]+[C2/(c-a)(c-b)]
=(ac+c2)/(c-a)(c-b)+[C2/(c-a)(c-b)]
=(ac+2c2)/((c-a)(c-b))
=(ac+2c2)/(ac+2c2)
=1