设A.B是方程(lgx)^2-lgx^2-2=0的两个根.求logAB(注A是下标.B是上标.+logBA(注B是下标.A是上标)=?

因为A，B是方程的(lgx)^2-lgx^2-2=0的两个根，把lgx看成是未知数，设y=lgx，那么lgA,lgB是方程y^2-2lgx-2=0的两根，由韦达定理可以得到lgA+lgB=2,(lgA)*(lgB)=-2,所以有lgA/lgB+lgB/lgA=[(lgA+lgB)^2-2lgAlgB]/lgAlgB=-4,而lgAB=lgB/lgA,lgBA=lgA/lgB,所以有lgAB+lgBA=-4.

-4

A.B是方程(lgx)^2-lgx^2-2=0的两个根
(lgx)^2-2lgx-2=0

lgA+lgB=2
lgA*lgB=-2

logA(B)+logB(A)=lgB/lgA+lgA/lgB=[(lgB)^2+(lgA)^2]/lgAlgB
=[(lgA+lgB)^2-2lgAlgB]/(-2)
=[4+4]/(-2)
=-4

lgA+lgB=1
lgA*lgB=-2
logAB+logBA=lgB/lgA+lgA/lgB=lgB ^2+lgA ^2 /lgA *lgB
=(lgA+lgB)^2/lgA*lgB -2 =1/-2 -2 =-5/2