已知函数f(x)=-x+loga^1-x/1+x,则f(-1/5)+f(-1/4)+f(-1/3)+f(1/5)+f(1/4)+f(1/3)+没限制范围已知函数f(x)=-x+loga^1-x/1+x,则f(-1/5)+f(-1/4)+f(-1/3)+f(1/5)+f(1/4)+f(1/3)+f(1/2)=?

Question

已知函数f(x)=-x+loga^1-x/1+x,则f(-1/5)+f(-1/4)+f(-1/3)+f(1/5)+f(1/4)+f(1/3)+
没限制范围 已知函数f(x)=-x+loga^1-x/1+x,则f(-1/5)+f(-1/4)+f(-1/3)+f(1/5)+f(1/4)+f(1/3)+f(1/2)=?

Answer 1

说实话!你写的这道题有歧义啊!就是你写的这道题不规范.可以有多种看法!也是,在电脑上不容易写明白作业本上的方程式!
不过,我做了多年的题,有点经验了!
f(-1/5)=1/5+loga(3/2)
f(-1/4)=1/4+loga(5/3)
f(-1/3)=1/3=loga2
f(1/5)=-1/5+loga(2/3)
f(1/4)=-1/4+loga(3/5)
f(1/3)=-1/3+loga(1/2)
f(-1/5)+f(-1/4)+f(-1/3)+f(1/5)+f(1/4)+f(1/3)=loga(3/2)+loga(5/3)+loga2+loga(2/3)+loga(3/5)+loga(1/2)
loga(3/2)+loga(2/3)=loga((3/2)x(2/3))=loga1=0
loga(5/3)+loga(3/5)=loga1=0
loga2+loga(1/2)=0
所以f(-1/5)+f(-1/4)+f(-1/3)+f(1/5)+f(1/4)+f(1/3)=loga(3/2)+loga(5/3)+loga2+loga(2/3)+loga(3/5)+loga(1/2)=0
f(-1/5)+f(-1/4)+f(-1/3)+f(1/5)+f(1/4)+f(1/3)+f(1/2)=f(1/2）=-1/2+loga(1/3)