已知:a-b=2,b-c=3,则a2+b2+c2-ab-bc-ca=_.

问题描述:

已知:a-b=2,b-c=3,则a2+b2+c2-ab-bc-ca=______.

∵a-b=2①,b-c=3②,
∴①+②得:a-c=5,
∴a2+b2+c2-ab-bc-ca=

1
2
(2a2+2b2+2c2-2ab-2bc-2ca)
=
1
2
[(a2-2ab+b2)+(a2-2ac+c2)+(b2-2bc+c2)]
=
1
2
[(a-b)2+(a-c)2+(b-c)2]
=
1
2
×[22+52+32]
=
1
2
×38
=19.
故答案为:19.