已知有理数x,y,z满足x+y-1+z-2=1/2(x+y+z),那么(x-yz)2的值为 _ .

问题描述:

已知有理数x,y,z满足

x
+
y-1
+
z-2
=
1
2
(x+y+z),那么(x-yz)2的值为 ___ .

将题中等式移项并将等号两边同乘以2得:x-2

x
+y-2
y-1
+z-2
z-2
=0
配方得(x-2
x
+1)+(y-1-2
y-1
+1)+(z-2-2
z-2
+1)=0

(
x
-1)
2
+(
y-1
-1)
2
+(
z-2
-1)
2
=0

x
=1且
y-1
=1且
z-2
=1

解得  x=1   y-2   z=3
∴(x-yz)2=(1-2×3)2=25