由题意可知:a-b=-1,b-c=-1,a-c=-2,
所求式=二分之一(2a^2+2b^2+2c^2-2ab-2bc-2ca)
=二分之一[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]
=二分之一[(a-b)^2+(b-c)^2+(a-c)^2]
=二分之一[(-1)^2+(-1)^2+(-2)^2]
=二分之4.
=2
由题意可知:a-b=-1,b-c=-1,a-c=-2,
所求式=二分之一(2a^2+2b^2+2c^2-2ab-2bc-2ca)
=二分之一[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]
=二分之一[(a-b)^2+(b-c)^2+(a-c)^2]
=二分之一[(-1)^2+(-1)^2+(-2)^2]
=二分之4.
=2