题目有误!若改成f(x/y)=f(x)-f(y)则可解.
(1)f(1)=f(1/1)=f(1)-f(1)=0
f(2)=f(1/(1/2))=f(1)-f(1/2)=0-1=-1
证明:任取x1<x2∈R+,则x2/x1>1∴f(x2)-f(x1)=f(x2/x1)<0
从而f(x)在定义域上单调递减
(2)f(x)+f(5-x)≥-2
<=>f(x)+f(5-x)≥2f(2)
<=>f(x)-f(2)≥f(2)-f(5-x)
<=>f(x/2)≥f[2/(5-x)]
<=>x/2≤2/(5-x)
<=>(x-4)(x-1)/2(x-5)≤0
<=>x≤1或4≤x<5
∴不等式解集为(-∞,1]∪[4,5)