设x1，x2∈（0，+∞）且x1＜x2∵f(x1)＝2x1−x1，f(x2)＝2x2−x2…2分∴f(x1)−f(x2)＝2x1−2x2+x2−x1=2(x2−x1)x1x2+x2−x1=(x2−x1)(2x1x2+1)…8分又∵0＜x1＜x2，∴x2−x1＞0，2x1x2+1＞0∴(x2−x1)(2x1x2+1)＞0...