(lgx)^2-lgx-2=0求logx1(x2)+logx2(x1)要过程

(lgx)^2-lgx-2=0
(lgx-2)(lgx+1)=0
lgx=2或lgx=-1
logx1(x2)+logx2(x1)=
=lgx1/lgx2+lgx2/lgx1
=2/(-1)+(-1)/2
=-2-1/2
=-5/2

设t=lgx
t^2-t-2=0
lg(x1)=t1=2 lg(x2)=t2=-1
logx1(x2)=lg(x2)/lg(x1)
所以求t2/t1+t1/t2=-2-0.5=-2.5

令t=lgx
t^2-t-2=0,t=2或-1,
则x1=10^(-1),x2=10^2
log10^(-1)(10^2)+log10^2(10^(-1))=-2log10(10)-1/2log10(10)
=-2-1/2
=-5/2