f（0+0）=f（0）+f(0),f(0)=0
f '(x)=lim_(y->0) [f(x+y)-f(x)]/y=lim_(y->0) [f(y)-f(0)]/y=f '(0)=1
f(x)=x+c
f(x+y)=f(x)+f(y),
x+y+c=x+c+y+c
c=0
所以
f(x)=x