已知x/2=y/3=z/4,求(x+y-2x)/(2x-y)的值
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已知x/2=y/3=z/4,求(x+y-2x)/(2x-y)的值
已知x/2=y/3=z/4,求(x+y-2x)/(2x-y)的值
已知x/2=y/3=z/4,求(x+y-2x)/(2x-y)的值
设x/2=y/3=z/4=k
则x=2k,
y=3k
z=4k.
(x+y-2x)/(2x-y)=(2k+3k-4k)/(4k-3k)
=1
由
x/2=y/3=z/4
得x=2y/3代入式子
(x+y-2x)/(2x-y)
(2y/3+y-2*2y/3)/(2*2y/3-y)=(y/3)/(y/3)=1
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