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
来源:学生作业学帮网 编辑:学帮网 时间:2024/05/30 02:36:39
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
a^lga*b^lgb*c^lgc=1求abc
已知a>1.b>1.c>1,且lga+lgb=1.求证:logaC+logbC>4lgc
如果lgx=lga+3lgb-5lgc,那么x,a,b,c的关系式为
若lgA=a,lgB=b,lgC=c,则(lgA^2)*[lg(B/2]*(lg√C)=?
已知(lgc/a)^2=4lga/b.lgb/c,则〡,b,c成A 等比数列 B 等差数列C 常数列 D 以上都不是
已知a>1,b>1,c>1且lga+lgb=1求证loga(a底数)c+logb(b底数)>=4lgc
已知M=A+B+C,为什么lgM=lgA+lgB+lgClgM/lgN,lg可以约掉变成M/N吗?如果M=abc,那么为什么lgM=lga+lgb+lgc?
已知a,b>1,0<c<1,且lga+lgb=1,求证:logac+logbc≤1lgc
若lga 、lgb、lgc成等差数列,则( )A.b=a+c/2 B.b=1/2(lga+lgc) C.a、b、c成等比数列 D.a、b、c成等差数列
已知lga,lgb,lgc与lga-lg2b,lg2b-lg3c,lg3c-lga依次成等差数列,求a,b,c之比
一道简单的高中数学选择题/.若lga,lgb,lgc成等差数列,则()A.b=(a+C)/2 B.b=(lga+lgc)/2 C.b=根号ac D.b=正负根号ac要过程喔~
lga-lgb=?(lga)/(lgb)=?lg(a/b)=?
已知a,b,c是不全相等的正数,求证:lga+lgb+lgc
lga-lgb=lg( a / b)lga+lgb=lg( a X b)lga X lgb=?lga / lgb=?
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
在直角三角形ABC中,角C=90°,边长a,b,c满足arcsin(1/a)+arcsin(1/b)=pai/2,求证;lgc=lga+lgb
RT三角形ABC中,角C=90度,边长a,b,c且arcsin1/a+arcsin1/b=π/2,求证:lgc=lga+lgb
RT三角形ABC中,角C=90度,边长a,b,c且arcsin1/a+arcsin1/b=π/2,求证:lgc=lga+lgb详细过程