Question

若|a-b-1|+(b-2a+c)²+|2c-b|=0求abc的值

Answer 1

因为绝对值和平方都是大于等于0的值
因此,a-b-1=0
(b-2a+c)^2=0
2c-b=0
解之：a=-3
b=-4
c=-2

追问

b为什么＝2c

追答

|2c-b|
=0
所以b=2c

Answer 2

由题可知：a-b-1=0, b+2a+c=0, 2c-b=0, 解得a=3, b=2,c=1

Answer 3

Answer 4

|a-b-1|+(b-2a+c)²+|2c-b|=0
=>
a-b-1=0 (1) and
b-2a+c=0 (2) and
2c-b=0
b/2=c = k
b=2k , c=k

from (2)

2k-2a+ k =0
a= (3/2)k

from (1)
a-b-1=0
(3/2)k - 2k -1=0
-(1/2)k -1=0
k=-2

a=(3/2)k = -3
b=2k=-4
c=k =-2

Answer 5

∵丨a-b-1丨+(b-2a+c)