答案:0
设1/2010=n,则-1/2010=-n
f(1/2010)+f(-1/2010)
=f(n)+f(-n)
=-n+log2 1-n/1+n+n+log2 1+n/1-n
=log2 [(1-n/1+n )(1+n/1-n)]
=log2 1
=0
答案:0
设1/2010=n,则-1/2010=-n
f(1/2010)+f(-1/2010)
=f(n)+f(-n)
=-n+log2 1-n/1+n+n+log2 1+n/1-n
=log2 [(1-n/1+n )(1+n/1-n)]
=log2 1
=0