下面就证明不等式问题经常出现的错误分类辨析如下,供大家参考.一、表述不规范【例1】设a≥3,则a - a -1 a -2- a -3.错解:由a - a -1 a -2- a -3,得a + a -3 a -1+ a -2.两边平方,得a(a -3) (a -1)(a -2) ,两边再平方,得a(a -3) (a -1)(a -2) ,即02,故原不等式得证.辨 The following is to prove that the inaccuracies of the inequalities often appear in the wrong classification and analysis as follows for your reference. First, the expression is not standardized [Example 1] Let a ≥ 3, then a - a -1 a -2- a -3. False solution: by a - a -1 a -2- a -3, get a + a -3 a -1+ a -2. Both sides of the square, get a (a -3) (a -1) (a -2), re-squared on both sides , get a(a -3) (a -1)(a -2), that is, 02, so the original inequality was obtained.