设f(x)在（0，+无穷）上是增函数，且f(x*y)=f(x)+f(y),又f(2)=1，求使不等式f(x)+F(x-3)<=2成立的x取值范

f(0)=f(0*0)=f(0)+f(0)=2f(0)
f(0)=0
f(4)=f(2*2)=f(2)+f(2)=1+1=2
f(x)+F(x-3)<=2
f[x(x-3)]<=f(4)
设f(x)在（0，+无穷）上是增函数
0<x(x-3)<=4

0<x(x-3)
因为f(x)在（0，+无穷），所以，x>3

x(x-3)<=4
x^2-3x-4<=0
(x-4)(x+1)<=0
-1<=x<=4

所以，3<x<=4

f(x)+F(x-3)=f(x*(x-3))<=2=f(1)+f(1)=f(1*1) 又递增，所以
0<x*(x-3)<1
x>0,x-3>0