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
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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](/uploads/image/z/8948028-12-8.jpg?t=%E5%B7%B2%E7%9F%A5a%3DlgA%2Cb%3DlgB%2Cc%3DlgC%2C%E6%BB%A1%E8%B6%B3ABC%3D1.%E6%B1%82%E8%AF%81%EF%BC%9AA%5E%281%2Fb%2B1%2Fc%29%C2%B7B%5E%281%2Fc%2B1%2Fa%29%C2%B7C%5E%281%2Fa%2B1%2Fb%29%3D1%2F1000)
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
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
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
设t=A^(1/b+1/c)·B^(1/c+1/a)·C^(1/a+1/b),a+b+c=lgA+lgB+lgC=lgABC=0
则lgt=(1/b+1/c)lgA+l(1/c+1/a)gB+(1/a+1/b)lgC=(a/b+a/c)+(b/c+b/a)+(c/a+c/b)
=(b+c)/a+(a+c)/b+(a+b)/c=-1-1-1=-3,所以t=1/1000,
即A^(1/b+1/c)·B^(1/c+1/a)·C^(1/a+1/b)=1/1000