a²+b²+c²=1,所以2（a²+b²+c²）=2,
又a-b=0.6,b-c=0.6,所以俩式相加a-c=1.2,（a-b)^2+(b-c)^2+(a-c)^2= 2a²+2b²+2c²-2ab-2bc-2ac=0.36+0.36+1.44=2.16,所以2ab-2bc-2ac=2.16-2=0.16,即ab+bc+ca=—0.08^是啥？抱歉 是平方（a-b)^2+(b-c)^2+(a-c)^2= 2a²+2b²+2c²-2ab-2bc-2ac=0.36+0.36+1.44=2.16,0.36+0.36+1.44=2.16 咋得来的？a-b=0.6,b-c=0.6,俩式想加，就是a-c=1.2了，根据完全平方公式，（a-b)²=a²-2ab+b²,所以(a-b)²+(b-c)²+(a-c)²=2a²+2b²+2c²-2ab-2bc-2ac