Question

(c-a)^2-4(a-b)(b-c)=0 求证a+c=2b

Answer 1

(c-a)^2-4(a-b)(b-c)

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

=c^2+a^2+2ac-4ab+4*b^2-4bc

=(a+c)^2-4b(a+c)+4b^2

=[(a+c)-2b]^2

=0

所以,

(a+c)-2b=0

即,a+c=2b,得证