若a^2+2b^2+2c^2-2ab-2bc+4-4c=0,求a,b,c的值?

a^2+2b^2+2c^2-2ab-2bc+4-4c=a^2-2ab+b^2+b^2-2bc+c^2+c^2-4c+4=(a-b)^2+(b-c)^2+(c-2)^2
=0 所以 a=b=c=2