Question

已知a=1996x+1995,b=1996x+1996,c=1996x+1997,求a^2+b^2+c^2-ab-bc-ca的值

Answer 1

a^2+b^2+c^2-ab-bc-ca
=1/2(2a^2+2b^2+2c^2-2ab-2ac-2bc)
=1/2[(a-b)^2+(a-c)^2+(b-c)^2]
因为a=1996x+1995,b=1996x+1996,c=1996x+1997
a-b=-1
a-c=-2
b-c=-1
所以=1/2[(a-b)^2+(a-c)^2+(b-c)^2]
=1/2(1+4+1)
=3

Answer 2

a-b=1
c-a=2
c-b=1
所以有：
a^2+b^2+c^2-ab-bc-ca
=1/2(a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2)
=1/2[(a-b)^2+(b-c)^2+(a-c)^2]
=1/2(1+1+4)
=3

Answer 3

2(a^2+b^2+c^2-ab-bc-ca)
=2a^2+2b^2+2c^2-2ab-2bc-2ca
=a^2+b^2-2ab+a^2+c^2-2ca+b^2+c^2-2bc
=(a-b)^2+(a-c)^2+(b-c)^2
=1+4+1=6
a^2+b^2+c^2-ab-bc-ca=3
祝你好运

Answer 4

配方得，所求式=[(a-b)^2+(b-c)^2+(a-c)^2]/2
=(1+1+2^2)/2
=3

Answer 5

a^2+b^2+c^2-ab-bc-ca=（（a-b）^2+（b-c）^2+（c-a）^2)/2
=3

Answer 6

=1/2(a-b)^2+1/2(a-c)^2+1/2(b-c)^2
=1/2+2+1/2
=3