Question

已知a²+b²+2c²+2ac-2bc=0,求a+b的值

Answer 1

原式=(a+c)^2+(b-c)^2=0,因为(a+c)^2>=0,(b-c)^2>=0,所以a+c=0,b-c=0,a=-c,b=c,a+b=0

Answer 2

(a c)^2 (b-c)^2=0 得a=-c b=c 故a b=0

Answer 3

a²+b²+2c²+2ac-2bc=0

(a+c)^2+(b-c)^2=0
a+c+b-c=0
a+b=0

Answer 4

a²+b²+2c²+2ac-2bc=0
a²+c²+2ac+c²+b²-2bc=0
(a+c)²+(b-c)²=0
a+c=0且b-c=0
a+b=0