希望有人帮助a=123456789 b=123456785 c=123456783求 a2(a的平方）+b2+c2-ab-bc-ac的值

(A-b)^2+(b-c)^2+(a-b)(b-c)

A、B、C转化即可.

a2(a的平方）+b2+c2-ab-bc-ac=[(a-b)2+(a-c)2+(b-c)2]/2=28

a2+b2+c2-ab-bc-ac
=1/2*(2a2+2b2+2c2-2ab-2bc-2ac)
=1/2*[(a-b)^2+(a-c)^2+(b-c)^2]
=1/2*(16+36+4)
=28

∵a=123456789 b=123456785 c=123456783
∴a-b=4,a-c=6,b-c=2
a²+b²+c²-ab-bc-ac
=1/2(2a²+2b²+2c²-2ab-2bc-2ac)
=1/2(a²+b²+c²+a²+b²+c²-2ab-2bc-2ac)
=1/2(a²-2ab+b²+a²-2ac+c²+b²-2bc+c²)
=1/2[(a-b)²+(a-c)²+(b-c)²]
=1/2[(a-b)²+(a-c)²+(b-c)²]
=1/2(4²+6²+2²)
=1/2(16+36+4)
=1/2×56
=28

a2(a的平方）+b2+c2-ab-bc-ac=a2-ab+b2-bc+c2-ac=4a+2b-4c=246913586

1/2[(a-b)2+(b-c)2+(c-a)2]=a2(a的平方）+b2+c2-ab-bc-ac
会了吧