已知a=lgA,b=lgB,c=lgC,满足ABC=1.求证:A^(1/b+1/c)·B^(1/c+1/a)·C^(1/
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已知a=lgA,b=lgB,c=lgC,满足ABC=1.求证:A^(1/b+1/c)·B^(1/c+1/a)·C^(1/a+1/b)=1/1000
a+b+c=lgA+lgB+lgC=lgABC=lg1=0
lg(A^(1/b+1/c)·B^(1/a+1/c)·C^(1/a+1/b))=(1/b+1/c)lgA+(1/a+1/c)lgB+(1/a+1/b)lgC=a/b+a/c+b/a+b/c+c/a+c/b
a+b+c=0 ,c=-a-b,a=-b-c,b=-a-c
a/b+a/c+b/a+b/c+c/a+c/b
=(-b-c)/b+(-b-c)/c+(-a-c)/a
=-1-c/b-b/c-1-1-c/a+b/c+c/a+c/b
=-3
lg(A^(1/b+1/c)·B^(1/a+1/c)·C^(1/a+1/b))=-3
所以A^(1/b+1/c)·B^(1/a+1/c)·C^(1/a+1/b)=10^(-3)=1/1000
lg(A^(1/b+1/c)·B^(1/a+1/c)·C^(1/a+1/b))=(1/b+1/c)lgA+(1/a+1/c)lgB+(1/a+1/b)lgC=a/b+a/c+b/a+b/c+c/a+c/b
a+b+c=0 ,c=-a-b,a=-b-c,b=-a-c
a/b+a/c+b/a+b/c+c/a+c/b
=(-b-c)/b+(-b-c)/c+(-a-c)/a
=-1-c/b-b/c-1-1-c/a+b/c+c/a+c/b
=-3
lg(A^(1/b+1/c)·B^(1/a+1/c)·C^(1/a+1/b))=-3
所以A^(1/b+1/c)·B^(1/a+1/c)·C^(1/a+1/b)=10^(-3)=1/1000
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