数学高手看过来:关于对数运算,急急急急急急急急急急急急急...

2lg2=lg4
2lg2+lg3=lg12

1=lg10
1/2lg0.36=lg0.6
1/3lg8=lg2
1+1/2lg0.36+1/3lg8=lg12

综上： lg12除以lg12=1

2lg2+lg3
=lg2^2+lg3
=lg2^2*3
=lg12

1+2/1lg0.36+3/1lg8
=1+lg0.36^(2/1)+lg8^(3/1)
=lg10+lg0.6+lg2
=lg(10*0.6*2)
=lg12

所以原式=lg12/lg12=1

你这写的让人看不明白啊。再叙述清楚一些

你的题目应该是2lg2+lg3除以1+1/2lg0.36+1/3lg8吧
=(lg2^2+lg3)/(1+lg0.36^1/2+lg8^1/3)
=(lg4+lg3)/(lg10+lg0.6+lg2)
=lg(4*3)/lg(10*0.6*2)
=lg12/lg12
=1

（2lg2+lg3）/(1+2/1lg0.36+3/1lg8)
=(lg2^2+lg3)/[lg10+lg0.36^(2/1)+lg8^(3/1)]
=lg12/lg12
=1

(2lg2+lg3)/(1+2/1lg0.36+3/1lg8)=(2lg2+lg3)/(1+1/(lg6-1)+1/lg2)=(2lg2+lg3)/[(2lg2+lg3)/(lg2+lg3-1)*lg2)]=(lg2+lg3-1)*lg2