f(x)=(ex-1)/(ex+1)=(e^x+1-2)/(e^x+1)=1-2/(e^x+1)设x2>x1>0,则f(x2)-f(x1)=[1-2/(e^x2+1)]-[1-2/(e^x1+1)]=-2/(e^x2+1)+2/(e^x1+1)=所以函数f(x)在区间(0,+∞)上是增函数