已知函数g(x)=1-2x,f[g(x)]=(1-x^2)/x^2 (x不等于0),则f(x)等于?

令a=g(x)=1-2x
x=(1-a)/2
所以f(a)=(1-x^2)/x^2=1/x^2-1=1/[(1-a)^2/4]-1=4/(a^2-2a+1)-1
所以f(x)=4/(x^2-2x+1)-1=(-x^2+2x+3)/(x^2-2x+1)
令g(x)=1-2x=0,x=1/2
所以f(0)=(1-x^2)/x^2=1/x^2-1=1/(1/4)-1=3