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
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.