按照增函数的定义来证明嘛!
令x1＜x2∈(-1,1)
则,f(x1)-f(x2)=[x1/(x1²-1)]-[x2/(x2²-1)]
=[x1*(x2²-1)-x2*(x1²-1)]/[(x1²-1)(x2²-1)]
=(x1x2²-x1-x2x1²+x2)/[(x1²-1)(x2²-1)]
=[x1x2(x2-x1)+(x2-x1)]/[(x1²-1)(x2²-1)]
=[(x1x2+1)(x2-x1)]/[(x1²-1)(x2²-1)]
已知x1＜x2∈(-1,1)
则,x1x2∈(-1,1),那么：x1x2+1∈(0,2)＞0
x2-x1＞0
(x1²-1)＜0,(x2²-1)＜0
所以,f(x1)-f(x2)=[(+)*(+)]/[(-)*(-)]＞0
即,f(x1)＞f(x2)
所以,f(x)在(-1,1)上是减函数!