【答案】
分析:(1)先移項(xiàng),然后對(duì)不等式的左邊進(jìn)行因式分解;最后根據(jù)不等式的性質(zhì)性質(zhì)解答;
(2)對(duì)分母的符號(hào)進(jìn)行分類(lèi)討論:當(dāng)2-x<0時(shí),去分母時(shí),不等號(hào)的方向發(fā)生改變;當(dāng)分母2-x>0時(shí),去分母時(shí),不等號(hào)的方向不變.
解答:解:(1)由原不等式,得
(2x+1)(3x-2)≥0,
∴①2x+1≥0,3x-2≥0,
解得,x≥
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163400810025423/SYS201310221634008100254013_DA/0.png)
;
②2x+1≤0,3x-2≤0,
解得,x≤-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163400810025423/SYS201310221634008100254013_DA/1.png)
;
故原不等式的解集是:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163400810025423/SYS201310221634008100254013_DA/2.png)
;
(2)當(dāng)分母2-x>0,即x<2時(shí),由原不等式,得
3x-1≥2-x,
即4x≥3,
∴x≥
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163400810025423/SYS201310221634008100254013_DA/3.png)
,
∴原不等式的解集是:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163400810025423/SYS201310221634008100254013_DA/4.png)
≤x<2;
當(dāng)分母2-x<0即x>2時(shí),由原不等式,得
3x-1≤2-x,
∴4x≤3,
∴x≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163400810025423/SYS201310221634008100254013_DA/5.png)
,
∴原不等式的解集是空集;
綜上所述,原不等式的解集是:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163400810025423/SYS201310221634008100254013_DA/6.png)
≤x<2.
點(diǎn)評(píng):本題考查了一元二次不等式的解法.解一元二次不等式的步驟是:①把二次項(xiàng)系數(shù)化為正數(shù);②解對(duì)應(yīng)的一元二次方程;③根據(jù)方程的根,結(jié)合不等號(hào)方向,得出不等式的解集.