函数fx满足关系f(xy)=f(x)+f(y) 求f(1/3)+f(1/2)+f(1)+f(2)+f(3)

函数fx满足关系f(xy)=f(x)+f(y) 求f(1/3)+f(1/2)+f(1)+f(2)+f(3)

答案是 0 由f(xy)=f(x)+f(y),得f(1)=0,如f(1/4)=f(1)+f(1/4)推出放f(1)=0
所以f(1/3)+f(1/2)+f(1)+f(2)+f(3)=f(1/3)+f(3)+f(1/2)+f(2)+f(1)
=f(1/3x3)+f(1/2x2)+f(1)
=f(1)+f(1)+f(1)
=0+0+0
=0

f(1/3)+f(1/2)+f(1)+f(2)+f(3)=f(1/6)+f(1)+f(6)=f(1/6)+f(6)=f(1)
f(1)=f(1)+f(1)=2f(1)
f(1)=0