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...
来源:百度知道 编辑:UC知道 时间:2024/07/02 11:28:46
2lg2+lg3除以1+2/1lg0.36+3/1lg8的运算过程及结果
(答对有加分哦!!!!!!!!!!!!)
...再加一道:lg4+lg9+2*根号下(lg6)^2-lg36+1
拜托啦~~~~~~`
(答对有加分哦!!!!!!!!!!!!)
...再加一道:lg4+lg9+2*根号下(lg6)^2-lg36+1
拜托啦~~~~~~`
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