因为a不等于b,所以lga不等于lgb
y由lg^2a+lg6·lga+lg2·lg3=0,(1)
lg^2b+lg6·lgb+lg2·lg3 .(2)
(1)-(2),得lg^2a-lg^2b+lg6(lga-lgb)=0
所以(lga-lgb)(lga+lgb)+lg6(lga-lgb)=0
而lga不等于lgb,即lga-lgb不等于0
所以lga+lgb+lg6=0
ab=1/6