已知F（x）在其定义域R+上为增函数,f(2)=1,f(xy)=f(x)+f(y),解不等式f(x)+f(x-2)

f(2)+f(2)=f(4)=1+1=2
f(2)+f(4)=f(8)=1+2=3
f(x)+f(x-2)=f[x(x-2)]=f(x^2-2x)
f(x)+f(x-2)