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已知x/(a+2b+c)=y/(a-c)=z/(a-2b=c)(abcxyz=0)求证a/(x+2y+z)=b/(x-z)=c/(x-2y+z)
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已知x/(a+2b+c)=y/(a-c)=z/(a-2b=c) (abcxyz=0) 求证a/(x+2y+z)=b/(x-z)=c/(x-2y+z)
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答案和解析
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)
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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