已知(z-x)的平方-4(x-y)(y-z)=0,求证x-2y+z=0
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![已知(z-x)的平方-4(x-y)(y-z)=0,求证x-2y+z=0](/uploads/image/z/5183605-37-5.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%88z-x%EF%BC%89%E7%9A%84%E5%B9%B3%E6%96%B9-4%EF%BC%88x-y%EF%BC%89%EF%BC%88y-z%EF%BC%89%3D0%2C%E6%B1%82%E8%AF%81x-2y%2Bz%3D0)
已知(z-x)的平方-4(x-y)(y-z)=0,求证x-2y+z=0
已知(z-x)的平方-4(x-y)(y-z)=0,求证x-2y+z=0
已知(z-x)的平方-4(x-y)(y-z)=0,求证x-2y+z=0
已知:
(z-x)^2-4(x-y)(y-z)=0
以下递推:
(x-z)^2-4(x-y)(y-z)=0
((x-y)+(y-z))^2-4(x-y)(y-z)=0
(x-y)^2+(y-z)^2+2(x-y)(y-z)-4(x-y)(x-z)=0
(x-y)^2+(y-z)^2-2(x-y)(y-z)=0
((x-y)-(y-z))^2=0
(x-2y+z)^2=0
x-2y+z=0.