(1)0
f(-2)=log2(1/2)+log2(2)=log2(1)-log(2)+log(2)=0
(2)0
猜想:f(x)+f(-x-1)=0
证明:假设x>0
f(-x-1)=log2[(-x-1+1)/(-x-1)]=log2[(-x)/(-x-1)]=log2(x)-log2(x+1)
f(x)=log2(x+1)-log2(x)
所以.f(x)+f(-x-1)=0
(1)0
f(-2)=log2(1/2)+log2(2)=log2(1)-log(2)+log(2)=0
(2)0
猜想:f(x)+f(-x-1)=0
证明:假设x>0
f(-x-1)=log2[(-x-1+1)/(-x-1)]=log2[(-x)/(-x-1)]=log2(x)-log2(x+1)
f(x)=log2(x+1)-log2(x)
所以.f(x)+f(-x-1)=0