一到高一的数学题目

m*n=1
由f(n)=f(m)得|log2m|=|log2n|,因为m<n,则log2m<log2n,故得 log2m=-log2n,
log2m+log2n=0,log2(m*n)=0,m*n=1,m=1/n
再由f(n)=2*f((m+n)/2),得|log2n|=2|log2((m+n)/2)|=2|log2((1+n^2)/(2n))|
(1+n^2)/(2n)>=1,|log2((1+n^2)/(2n))|=log2((1+n^2)/(2n)),|log2n|=log2n,于是得log2n=2log2((1+n^2)/(2n)),n=((1+n^2)/(2n))^2,

log2m|=|log2n|=|log(((m+n)/2)^2)|=|log((m^2+n^2)/4+mn/2)|
m<n,则log2m!=(不等于)log2n
=>log2m=-log2n=>2m=1/(2n)=>m*n=1/4

tnnd,我最讨厌问我数学题目的人。

|log2m|=|log2n|=|log(m+n)^2)|=|log((m^2+n^2)/4+mn/2)|
m<n,则log2m!=(不等于)log2n
=>log2m=-log2n=>2m=1/(2n)=>m*n=1/4

然后不知道你是log2x还是log以2为底x
==帮你解