函数f(x)对任意实数a,b,都有f(ab)=f(a)+f(b)成立(1)令a=0,b=0那么有f(ab)=f(a)+f(b)f(0*0)=f(0)+f(0)f(0)=f(0)+f(0)故f(0)=0(2)令a=1,b=1那么有f(ab)=f(a)+f(b)f(1*1)=f(1)+f(1)f(1)=f(1)+f(1)故f(1)=0呵呵~这样解答...f(0)=f(0)+f(0)等式左右两边同时减去f(0)可得0=f(0)f(1)=f(1)+f(1)等式左右两边同时减去f(1)可得0=f(1)呵呵