有当x>=1时,1/f'（x）＞1得：当x>=1时,0＜f'（x）＜1设x1＜x2＜1,则2-x1＞2-x2＞1f（x+1）为奇函数,即f（x+1）=-f（-x+1）,得f（x）=-f（2-x）,f（1）=0于是f（x1）-f（x2）=-f（2-x1）+f（2-x2）＜0于是当x∈（-无...