证明：（1）设任取x1，x2∈[3，5]且x1＜x2f(x1)−f(x2)＝x1−1x1+2−x2−1x2+2＝3(x1−x2)(x1+2)(x2+2)∵3≤x1＜x2≤5∴x1-x2＜0，（x1+2）（x2+2）＞0∴f（x1）-f（x2）＜0即f（x1）＜f（x2）∴f（x）在[3，5]上为...