更新时间2019-12-12 01:48:58
a2 + b2 + c2 - ab - bc - ac
= (1/2)[ ( a - b )^2 + ( a - c )^2 + ( b - c )^2 ]
= (1/2)[ 1 + 4 + 1 ]
= 3
a^2+b^2+c^2-ab-bc-ca
=1/2[2a^2+2b^2+2c^2-2ab-2bc-2ca]
=1/2[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)]
=1/2[(a-b)^2+(b-c)^2+(c-a)^2]
=1/2[(2018-2019)^2+(2019-2020)^2+(2020-2018)^2]
=1/2(1+1+4)
=3
a²+b²+c²-ab-bc-ac
=(2a²+2b²+2c²-2ab-2bc-2ac)÷2
=(a²+b²-2ab+a²+c²-2ac+b²+c²-2bc)÷2
=[(a-b)²+(a-c)²+(b-c)²]÷2
将a=2018，b=2019，c=2020代入上式：
（1+4+1）÷2
=6÷2
=3
a2+b2+c2-ab-bc-ca=12[2a2+2b2+2c2-2ab-2bc-2ca]=12[（a2-2ab+b2）+（b2-2bc+c2）+（c2-2ac+a2）]=12[（a-b）2+（b-c）2+（c-a）2]=12*{1+1+4}=72
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