Question

[2*根号x]+[根号（y-1）]+[根号（z-2）]=x+y+z ,求x,y,z的值？

Answer 1

通过移项，可得
x - 2√x + y - √(y-1) + z- √(z-2) =0
配方
(x - 2√x + 1) + (y-1 - √(y-1) +1/4) +(z-2 - √(z-2) + 1/4) = 1/2 -2
即
(√x -1)^2 + (√(y-1) - 1/2)^2 + (√(z-2) - 1/2)^2 = -3/2 < 0，无解

Answer 2

本题中括号里应该都有一个2
x-[2*根号x]+1+(y-1)-[2根号（y-1）]+1+(z-2)-[2根号（z-2）]+1=0
[根号x-1]²+[根号(y-1)-1]²+[根号(z-2)-1]²=0
满足上式，只需
根号x-1=0
根号(y-1)-1=0
根号(z-2)-1=0
解得x=1
y-1=1 y=2
z-2=1 z=3

追问

中括号内没有2？

Answer 3

如果题目是 [2*根号x]+[根号（y-1）]+[根号（z-1/2）]=x+y+z 则：

x - 2根号x +1 + (y-1) -根号(y-1)+1/4 + (z-1/2)-根号(z-1/2)+1/4 = 0
(根号x-1)^2 + (根号（y-1）-1/2)^2 +(根号（z-1/2）-1/2)^2 = 0
根号x-1 = 0, 根号（y-1）-1/2 = 0,根号（z-1/2）-1/2 =0
所以：
x = 1
y = 5/4
z = 3/4