已知a=1996x+1995,b=1996x+1996,c=1996x+1997,求a^2+b^2+

网友回答

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======以下答案可供参考======
供参考答案1:
a-b=1c-a=2c-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供参考答案2:
a^2+b^2+c^2-ab-bc-ca
=（（a-b）^2+（b-c）^2+（c-a）^2)/2=3供参考答案3:
=1/2(a-b)^2+1/2(a-c)^2+1/2(b-c)^2
=1/2+2+1/2
=3供参考答案4:
配方得，所求式=[(a-b)^2+(b-c)^2+(a-c)^2]/2
=(1+1+2^2)/2
=3供参考答案5:
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