设x/(a+2b+c)=y/(a-c)=z/(a-2b+c) = k,（显然k≠0）则x = (a+2b+c)ky = (a-c)kz = (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...