请问lg5*lg8000+[lg2底的根号三次方]的平方 尀 lg600-1尀2lg0.036-1尀2lg0.1 如何计算?

2024-12-12 01:53:53
推荐回答(1个)
回答1:

[lg5*lg8000+(lg2^根三)^2]/(lg600-1/2lg0.036-1/2lg0.1)
=[(lg5*(lg8+3)+(√3*lg2)^2]/[lg6+2-1/2lg(0.036*0.1)]
=[3lg5*lg2+3lg5+3*(lg2)^2]/[lg6+2-lg√(0.0036)]
=[3lg2(lg5+lg2)+3lg5]/[lg6+2-lg0.06]
=(3lg2+3lg5)/[lg(6/0.06)+2]
=3(lg5+lg2)/(lg100+2)
=3/(2+2)
=3/4