Question

{√[(lg3)^2-lg9+1] *(lg√27+lg8-lg√1000)}/(lg0.3*lg1.2)

Answer 1

√[(lg3)^2-lg9+1]=√(lg3-1)^2=1-lg3

lg√27+lg8-lg√1000=3/2*lg3+3lg2-3/2=3/2(lg3+2lg2-1)

lg0.3*lg1.2=(lg3-1)(lg12-1)=(lg3-1)(lg3+2lg2-1)

原式=3/2*(1-lg3)(lg3+2lg2-1) / [(lg3-1)(lg3+2lg2-1)]

=-3/2