证明 (10 21:28:12)

函数f (x)=x/(x^2-1)在区间（-1,1）上单调递减.
证明：设-1＜a＜b＜1
f(a)-f(b)=a/(a^2-1)-b/(b^2-1)
=(ab^2-a-ba^2+b)/(a^2-1)(b^2-1)
=(ab+1)(b-a)/(a^2-1)(b^2-1)
∵-1＜x＜1∴ ab+1＞0 a^2-1＜0 b-a>0
∴ (ab+1)(b-a)＞0
(a^2-1)(b^2-1)＞0
∴f(a)＞f(b) ∵a＜b
∴ 函数f (x)=x/x^2-1在区间（-1,1）上单调递减.