Question

若（a-c)^2-4（a-b）（b-c）=0，求a-2b+c，求过程（两种方法）

Answer 1

a-2b+c=0
令a-b=m c-b=n 则a-c=m-n a-2b+c=m+n
(a-c)^2-4(a-b)(b-c)=(m-n)^2+4mn=0
即m^2-2mn+n^2+4mn=m^2+2mn+n^2=(m-n)^2=0
m-n=0
即a-2b+c=0

Answer 2

当 a=b时a=c,a-2b+c=0
假设方程（a-b)x^2+(a-c)x+(b-c)=0=[(a-b)x+(b-c)](x+1)
方程有解-1，(b-c)/(a-b)
有（a-c）的平方-4（a-b）（b-c）=0可知两根相等即，
(b-c)/(a-b)=-1
a-2b+c=0

Answer 3

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

Answer 4

(a-c)^2-4(a-b)(b-c) =0
[(a-b)+(b-c)]^2-4(a-b)(b-c) =0
(a-b)^2+2(a-b)(b-c)+(b-c)^2-2(a-b)(b-c)=0
(a-b)^2+(b-c)^2-2(a-b)(b-c)=0
[(a-b)-(b-c)]^2=0
(a+c-2b)^2=0
(a-2b+c)^2=0