问题 填空题
若有理数x,y,z满足
x
+
y-1
+
z-2
=
1
2
(x+y+z),则(x-yz)3的值为______.
答案

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)3=-125.

故答案为:-125.

单项选择题
单项选择题