若(lgx)^2-lgx^2-2=0的两根为a,b,则loga b+logb a的值是

(lgx)^2-lgx^2-2=0,则(lgx)^2-2lgx-2=0,设y=lgx,则y²-2y-2=0,设两个根是y1,y2,
根据韦达定理,得y1+y2=2,y1*y2=-2
因为y1=lga,y2=lgb,则lga+lgb=2,lga*lgb=-2
loga b+logb a=lgb/lga+lga/lgb=lg²b/(lga*lgb)+lg²a/(lga*lgb)=(lg²a+lg²b)/(lga*lgb)
=[(lga+lgb)²-2lga*lgb]/(lga*lgb)=(2²+4)/(-2)=8/(-2)=-4

令lgX等于t，带入式子中得出两根为1/10;100，用换底公式化简式子，带入得-5/2

令lgx=t
则t^2-2t-2=0,t1+t2=2,t1t2=-2
所以loga b+logb a=lgb/lga+lga/lgb
=t2/t1+t1/t2
=(t1^2+t2^2)/t1t2
=[(t1+t2)^2-2t1t2]/t1t2=-4

a，b是方程（lgx）^2-lgx^2-2=0的两个根
即:
(lga)^2-2lga-2=0
(lgb)^2-2lgb-2=0
设m=lga,n=lgb
则:m^2-2m-2=0,n^2-2n-2=0
说明m,n是方程t^2-2t-2=0的二个根.
m+n=2,mn=-2
loga b+logb a=lgb/lga+lga/lgb=[(lgb)^2+(lga)^2]/(lga*lgb)
=[(m+n)^2-2mn]/(mn)
=[4-2*(-2)]/(-2)
=-4