已知a+b+c=0,则a2/(2a2+bc)+b2/(2b2+ac)+c2/(2c2+ab)的值为多少?
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已知a+b+c=0,则a2/(2a2+bc)+b2/(2b2+ac)+c2/(2c2+ab)的值为多少?
已知a+b+c=0,则a2/(2a2+bc)+b2/(2b2+ac)+c2/(2c2+ab)的值为多少?
已知a+b+c=0,则a2/(2a2+bc)+b2/(2b2+ac)+c2/(2c2+ab)的值为多少?
原式=a²/[2a²-b﹙a+b﹚]+b²/[2b²-c﹙b+c﹚]+c²/[2c²-a﹙a+c﹚]=a²/[﹙2a+b﹚﹙a-b﹚]+b²/[﹙2b+c﹚﹙b-c﹚]+c²/[﹙2c+a﹚﹙c-a﹚]=a²/[﹙a-c﹚﹙a-b﹚]+b²/[﹙b-a﹚﹙b-c﹚]+c²/[﹙c-b﹚﹙c-a﹚]=[-a²﹙b-c﹚-b²﹙c-a﹚-c²﹙a-b﹚]/[﹙a-b﹚﹙b-c﹚﹙c-a﹚]=[-a²﹙b-c﹚-bc﹙b-c﹚+a﹙b+c﹚﹙b-c﹚]/[﹙a-b﹚﹙b-c﹚﹙c-a﹚]=﹙b-c﹚[-a﹙a-b﹚+c﹙a-b﹚]/﹙a-b﹚﹙b-c﹚﹙c-a﹚=[﹙a-b﹚﹙b-c﹚﹙c-a﹚]/[﹙a-b﹚﹙b-c﹚﹙c-a﹚]=1
a2/(2a2+bc)+b2/(2b2+ac)+c2/(2c2+ab)=0
3/2
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