f(x)是定义在（0，正无穷）的增函数，f(x/y)=f(x)-f(y)
求f(1)的值
若f(6)=1 解不等式f(x+3)-f(1/x)<2
參考答案:(1)f(1)=f(1/1)=f(1)-f(1)=0
(2)f(x+3)-f(1/x)<2
=>f(x+3)-f(1)+f(x)<2
=>f(x+3)+f(x)<2
=>f(6)=1
=>f(x+3)-f(6)<f(6)-f(x)
=>f[(x+3)/6]<f(6/x)
=>f(x) 增函数
=>(x+3)/6<6/x
=>x^2+3x-36<0
=>解之即可。
