1/(1+log3 2)+1/(1+log2 3)=lg3/(lg2+lg3)+lg2/(lg2+lg3)=1解释一下这条式子是怎么转换的
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1/(1+log32)+1/(1+log23)=lg3/(lg2+lg3)+lg2/(lg2+lg3)=1解释一下这条式子是怎么转换的1/(1+log32)+1/(1+log23)=lg3/(lg2+
1/(1+log3 2)+1/(1+log2 3)=lg3/(lg2+lg3)+lg2/(lg2+lg3)=1解释一下这条式子是怎么转换的
1/(1+log3 2)+1/(1+log2 3)=lg3/(lg2+lg3)+lg2/(lg2+lg3)=1解释一下这条式子是怎么转换的
1/(1+log3 2)+1/(1+log2 3)=lg3/(lg2+lg3)+lg2/(lg2+lg3)=1解释一下这条式子是怎么转换的
因为log3 2 = lg2/lg3,log2 3 = lg3/lg2
所以1/(1+log3 2)+1/(1+log2 3)= 1/(1+lg2/lg3)+1/(1+lg3/lg2)
前一项分子分母同时乘以lg3,后一项分子分母同时乘以lg2
则1/(1+log3 2)+1/(1+log2 3)= 1/(1+lg2/lg3)+1/(1+lg3/lg2)
=lg3/(lg2+lg3)+lg2/(lg2+lg3)=(lg3+lg2)/(lg2+lg3) = 1
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log2 1-lg3×log3 2-lg5