设x1＜x2＜1，则f（x1）-f（x2）=1(x1−1)2-1(x2−1)2=(x1+x2−2)(x2−x1)(x1−1)2(x2−1)2 ∵x1＜x2＜1，∴x2-x1＞0，x1+x2＜2，x1+x2-2＜0∴f（x1）-f（x2）＜0，即f（x1）＜f（x2），∴函数f（x）=1(x−1)2...