已知(x-1)²+2丨y-z-2丨+(y-2x)²=0,则xyz=
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![已知(x-1)²+2丨y-z-2丨+(y-2x)²=0,则xyz=](/uploads/image/z/7084016-8-6.jpg?t=%E5%B7%B2%E7%9F%A5%28x-1%29%26%23178%3B%2B2%E4%B8%A8y-z-2%E4%B8%A8%2B%28y-2x%29%26%23178%3B%3D0%2C%E5%88%99xyz%3D)
已知(x-1)²+2丨y-z-2丨+(y-2x)²=0,则xyz=
已知(x-1)²+2丨y-z-2丨+(y-2x)²=0,则xyz=
已知(x-1)²+2丨y-z-2丨+(y-2x)²=0,则xyz=
∵(x-1)²+2丨y-z-2丨+(y-2x)²=0
∴(x-1)²≥0,2丨y-z-2丨≥0,(y-2x)²≥0
即x=1,y-z-2=0,y-2x=0
即x=1,y=2,z=0
∴xyz=1×2×0=0
x-1=0
y-z-2=0
y-2x=0
解得
x=1
y=2
z=0
所以
xyz=0
(x-1)²+2丨y-z-2丨+(y-2x)²=0
x-1=0 y-z-2=0 y-2x=0(非负数之和为0,各自为0)
x=1
y=2
z=0
xyz=0
祝你好运