(lgx)^2-lgx-2=0
a,b是方程的两个实数根
即lga,lgb是方程t^2-t-2=0的二个根.
韦达定理得:
lga+lgb=lgab=1
lga*lgb=-2
是否是求:loga b+logb a
=lgb/lga+lga/lgb
=[(lgb)^2+(lga)^2]/lga*lgb
=[(lga+lgb)^2-2lga*lgb]/(-2)
=(1+4)/(-2)
=-6/2 (lgx)^2-lgx-2=0
a,b是方程的两个实数根
即lga,lgb是方程t^2-t-2=0的二个根.
韦达定理得:
lga+lgb=lgab=1
lga*lgb=-2
是否是求:loga b+logb a
=lgb/lga+lga/lgb
=[(lgb)^2+(lga)^2]/lga*lgb
=[(lga+lgb)^2-2lga*lgb]/(-2)
=(1+4)/(-2)
=-6/2