若有理数x,y,z满足(x-1)²+(2x-y)的4次方+(x-1z)²=0,求x+y+z的值
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![若有理数x,y,z满足(x-1)²+(2x-y)的4次方+(x-1z)²=0,求x+y+z的值](/uploads/image/z/10001864-56-4.jpg?t=%E8%8B%A5%E6%9C%89%E7%90%86%E6%95%B0x%2Cy%2Cz%E6%BB%A1%E8%B6%B3%28x-1%29%26%23178%3B%EF%BC%8B%EF%BC%882x%EF%BC%8Dy%EF%BC%89%E7%9A%844%E6%AC%A1%E6%96%B9%EF%BC%8B%EF%BC%88x%EF%BC%8D1z%EF%BC%89%26%23178%3B%3D0%2C%E6%B1%82x%2By%2Bz%E7%9A%84%E5%80%BC)
若有理数x,y,z满足(x-1)²+(2x-y)的4次方+(x-1z)²=0,求x+y+z的值
若有理数x,y,z满足(x-1)²+(2x-y)的4次方+(x-1z)²=0,求x+y+z的值
若有理数x,y,z满足(x-1)²+(2x-y)的4次方+(x-1z)²=0,求x+y+z的值
要满足以上条件,则各个式子应该为O.
解方程组x-1=0;
2x-y=0;
x-z=0.
得x=1,y=2,z=1,所以x+y+z=4.
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x=1 y=2 z=1 x+y+z=4
(x-1)²+(2x-y)^4+(x-z)²=0
因为(x-1)²≥0 (2x-y)^4≥0 (x-z)²≥0
所以x-1=0 x=1
2x-y=0 y=2
x-z=0 z=1
x+y+z=1+2+1=4