【答案】
分析:分兩種情況考慮:(i)a等于0求出解集;(ii)a不為0時,由a小于1,再分a大于0小于1與a小于0兩種情況考慮,分別求出相應(yīng)的解集即可.
解答:解:(i)a=0時,不等式解集為x∈R且x≠2;
(ii)a≠0時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024913115493/SYS201311031030249131154017_DA/0.png)
<1?
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024913115493/SYS201311031030249131154017_DA/1.png)
>0?[(a-1)x+2](x-2)>0,
∵a<1,∴a-1<0,
∴化為(x-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024913115493/SYS201311031030249131154017_DA/2.png)
)(x-2)<0,
(a)當(dāng)0<a<1時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024913115493/SYS201311031030249131154017_DA/3.png)
>2,
∴不等式的解為2<x<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024913115493/SYS201311031030249131154017_DA/4.png)
;
(b)當(dāng)a<0時,1-a>1,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024913115493/SYS201311031030249131154017_DA/5.png)
<2,
∴不等式解為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024913115493/SYS201311031030249131154017_DA/6.png)
<x<2,
則當(dāng)0<a<1時,不等式解集為{x|2<x
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024913115493/SYS201311031030249131154017_DA/7.png)
};當(dāng)a<0時,不等式解集為{x|
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103024913115493/SYS201311031030249131154017_DA/8.png)
};當(dāng)a=0時,解集為{x∈R|x≠2}.
點評:此題考查了其他不等式的解法,利用了分類討論的思想,分類討論時注意考慮問題要全面,做到不重不漏.