若a,b,c为实数,关于x的方程2x2+2（a-c）x+(a-b)2+(b-c)2=0有两个相等的实数根,求证：a+c=

4(a-c)^2 -4*2*[(a-b)^2 +(b-c)^2]=0
(a-c)^2 =2*(a-b)^2 +2(b-c)^2
a^2-2ac +c^2 =2a^2 -4ab +2b^2 +2b^2-4bc+2c^2
a^2 +4b^2 +c^2 -4ab -4bc +2ac =0
(a-2b)^2 +2c*(a-2b) +c^2 =0
(a-2b +c)^2=0
a-2b+c =0
a+c=2