若x/(a+2b+c)=y/(a-c)=z/(a-2b+c),且a,b,c,x,y,z均不为0,求证a/(x+2y+z)
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若x/(a+2b+c)=y/(a-c)=z/(a-2b+c),且a,b,c,x,y,z均不为0,求证a/(x+2y+z)=b/(x-z)=c/(x-2y+z)
![若x/(a+2b+c)=y/(a-c)=z/(a-2b+c),且a,b,c,x,y,z均不为0,求证a/(x+2y+z)](/uploads/image/z/19714552-16-2.jpg?t=%E8%8B%A5x%2F%28a%2B2b%2Bc%29%3Dy%2F%28a-c%29%3Dz%2F%28a-2b%2Bc%29%2C%E4%B8%94a%2Cb%2Cc%2Cx%2Cy%2Cz%E5%9D%87%E4%B8%8D%E4%B8%BA0%2C%E6%B1%82%E8%AF%81a%2F%28x%2B2y%2Bz%29)
设x/(a+2b+c)=y/(a-c)=z/(a-2b+c) = k, (显然k≠0) 则 x = (a+2b+c)k y = (a-c)k z = (a-2b+c)k 于是(直接代入), a/(x+2y+z) = a/(4ak) = 1/(4k) b/(x-z) = b/(4bk) = 1/(4k) c/(x-2y+z) = c/(4ck) = 1/(4k) 综上所述,a/(x+2y+z)=b/(x-z)=c/(x-2y+z)
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