a,b,c,互不相等,a+b+c=0 则 a2/2a2+bc+ b2/2b2+ac + c2/2c2+ab=?

a+b+c=0,-(a+b)=c,-(b+c)=a,-(a+c)=b,代入：
a2/2a2+bc+ b2/2b2+ac + c2/2c2+ab
=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