已知函数f(x)=(2x)/(x^2+1)

f(x)=(2x)/(x^2+1)为减函数
设x1>x2>1
f(x1)-f(x2)=2x1/(x1^2+1)-2x2/(x2^2+1),
=[2x1(x2^2+1)-2x2(x1^2+1)]/(x1^2+1)(x2^2+1)
=[2x1x^2+2x1-2x2x1^2-2x2]/(x1^2+1)(x2^2+1)
=[2x1x2(x2-x1)-2(x2-x1)]/(x1^2+1)(x2^2+1)
=[2(x2-x1)(x1x2-1)]/(x1^2+1)(x2^2+1)
(x2-x1)0
所以f(x1)-f(x2)