已知:a+b+c=0,且ab≠0,试证明:[a2/(2a2+bc)]+[b2/(2b2+ac)]+[C2/(2c2+ab)]=1_数学解答

已知:a+b+c=0,且ab≠0,试证明:[a2/(2a2+bc)]+[b2/(2b2+ac)]+[C2/(2c2+ab)]=1

人气：231 ℃　时间：2019-10-26 11:00:06

解答

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...