1、a²+b²-4a+8b+20
= a²-4a+4+b²+8b+16
= (a-2)^2 + (b+4)^2
=0
所以 a= 2 b= -4
所以 a+b = -2
2、 a²+b²+c²-ab-bc-ca
乘以2后再除以2 = 0.5(a²-2ab+b²+a²-2ac+c²+b²-2bc+c²)
=0.5[(a-b)^2+(a-c)^2+(b-c)^2]
=0.5[1+4+1]
=3