证明:(a-b)²分之1+(b-c)²分之1+(c-a)²分之1=(a-b分之1+(b-c)分之1+(c-a)分之1)²
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证明:(a-b)²分之1+(b-c)²分之1+(c-a)²分之1=(a-b分之1+(b-c)分之1+(c-a)分之1)²
证明:(a-b)²分之1+(b-c)²分之1+(c-a)²分之1=(a-b分之1+(b-c)分之1+(c-a)分之1)²
证明:(a-b)²分之1+(b-c)²分之1+(c-a)²分之1=(a-b分之1+(b-c)分之1+(c-a)分之1)²
∵1/[﹙a-b﹚﹙b-c﹚]+1/[﹙c-a﹚﹙b-c﹚]+1/[﹙a-b﹚﹙c-a﹚]
=﹙c-a﹚/[﹙a-b﹚﹙b-c﹚﹙c-a﹚]+﹙a-b﹚/[﹙c-a﹚﹙b-c﹚﹙a-b﹚]+﹙b-c﹚/[﹙a-b﹚﹙c-a﹚﹙b-c﹚]
=[﹙c-a﹚+﹙a-b﹚+﹙b-c﹚]/[﹙a-b﹚﹙b-c﹚﹙c-a﹚]
=0
∴1/(a-b)²+1/(b-c)²+1/(c-a)²+2/[﹙a-b﹚﹙b-c﹚]+2/[﹙c-a﹚﹙b-c﹚]+2/[﹙a-b﹚﹙c-a﹚]=1/(a-b)²+1/(b-c)²+1/(c-a)²
即1/(a-b)²+1/(b-c)²+1/(c-a)²=(a-b分之1+(b-c)分之1+(c-a)分之1)²
好多括号啊………………………………………………
小学一年级
好那个啊.......