数学题解出给高分

（1）．
因为 4(a-b)(b-c)-(a-c)^2 = 0

即 (a-c)^2 - 4(a-b)(b-c) = 0

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

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

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

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

a + c - 2b = 0

所以 a+c=2b

（2）因为 m + 2n - 1 = 0

所以 m + 2n = 1

则(m +2 n)^2 - 2n - m = 1 - 2n - m = 1 - (m + 2n) = 0

1)
4(a-b)(b-c)-(a-c)^2=0,
4ab-4ac-4b^2+4bc-a^2+2ac-c^2=0,
4ab-2ac-4b^2+4bc-a^2-c^2=0,
(a+c-2b)^2=0,
a+c=2b

2)(m+2n)^2-2n-m
=(m+2n)^2-(2n+m)
=1^2-1
=0

1)
4(a-b)(b-c)-(a-c)^2=0,
4ab-4ac-4b^2+4bc-a^2+2ac-c^2=0,
4ab-2ac-4b^2+4bc-a^2-c^2=0,
(a + c)^2 - 4b(a + c) + 4b^2 = 0
这一步最重要是完全平方形式
(a+c-2b)^2=0,
a+c=2b

2)因为 m + 2n - 1 = 0
所以 m + 2n = 1
代入就好了
(m+2n)^2-2n-m
=(m+2n)^2-(2n+m)
=1^2-1
=0

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