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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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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