Simone若函数f(x)在其定义域(0,+∞)上为增函数，f(2)=1,f(xy)=f(x)+f(y)，则不等式f(x)+f(x-3)≤2的解集是？
的有关信息介绍如下：
f(2) = f(1*2) = f(1) + f(2) f(1) = 0f(4) = f(2*2) = f(2) + f(2) = 1 + 1 = 2f(x)在其定义域(0,+∞)上为增函数, x=4时，f(x) + f(x-3) = f(4) + f(1) = 2 + 0 = 2所以f(x)+f(x-3)≤2，x≤4， 要使f(x)和f(x-3)都有定义，x-3>0, x>3解为3
