lgx+lgy+lgz
=lgxyz
=0=lg1
xyz=1
所以xz=1/y xy=1/z yz=1/x
则(xz)^(1/lgy)*(xy)^(1/lgz)*(yz)^(1/lgx)
=(1/y)^(1/lgy)*(1/z)^(1/lgz)*(1/x)^(1/lgx)
=(1/y)^(-lgy)*(1/z)^(-lgz)*(1/x)^(-lgx)
=y^(lgy)*z^(lgz)*x^(lgx)
因a^(loga b)=b logb a=1/loga b(基本公式)
a^(logb a)=a^[1/loga b]=a^[(loga b)^(-1)]=[a^(loga b)]^(-1)=b^(-1)=1/b
原式=(1/10)*(1/10)*(1/10)=1/1000
就得出结果了.(1/y)^(1/lgy)*(1/z)^(1/lgz)*(1/x)^(1/lgx)=(1/y)^(-lgy)*(1/z)^(-lgz)*(1/x)^(-lgx)这两个是怎么等起来的？哦，电脑里看错了。笔算不会有的(1/y)^(1/lgy)*(1/z)^(1/lgz)*(1/x)^(1/lgx)=(1/y)^(lg10/lgy)*(1/z)^(lg10/lgz)*(1/x)^(lg10/lgx)=(y^-1)^logy 10*(z^-1)^logz 10* (x^-1)^logx 10=(y^logy 10)^(-1)*(z^logz 10)^(-1)*(x^logx 10)^(-1)