已知a(y-z)+b(z-x)+c(x-y)=0,求证:(x-y)/(a-b)=(y-z)/(b-c)=(z-x)/(c-a)
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![已知a(y-z)+b(z-x)+c(x-y)=0,求证:(x-y)/(a-b)=(y-z)/(b-c)=(z-x)/(c-a)](/uploads/image/z/11957125-13-5.jpg?t=%E5%B7%B2%E7%9F%A5a%28y-z%29%2Bb%28z-x%29%2Bc%28x-y%29%3D0%2C%E6%B1%82%E8%AF%81%3A%28x-y%29%2F%28a-b%29%3D%28y-z%29%2F%28b-c%29%3D%28z-x%29%2F%28c-a%29)
已知a(y-z)+b(z-x)+c(x-y)=0,求证:(x-y)/(a-b)=(y-z)/(b-c)=(z-x)/(c-a)
已知a(y-z)+b(z-x)+c(x-y)=0,求证:(x-y)/(a-b)=(y-z)/(b-c)=(z-x)/(c-a)
已知a(y-z)+b(z-x)+c(x-y)=0,求证:(x-y)/(a-b)=(y-z)/(b-c)=(z-x)/(c-a)
a(y-z)+b(z-x)+c(x-y)=0
可化简成
a(y-z)+b((z-y)+(y-x))+c(x-y)=0
分解因式可得前边部分,然后再分解y-z或者x-y即可
令a(y-z)=0 b(z-x)=0 c(x-y)=0
所以X=Y=Z
就可以证明
(x-y)/(a-b)=(y-z)/(b-c)=(z-x)/(c-a)=0