【答案】
分析:(1)原式第一項(xiàng)化為最簡(jiǎn)二次根式,第二項(xiàng)利用負(fù)指數(shù)冪法則計(jì)算,第三項(xiàng)利用絕對(duì)值的代數(shù)意義化簡(jiǎn),最后一項(xiàng)利用特殊角的三角函數(shù)值化簡(jiǎn),計(jì)算即可得到結(jié)果;
(2)分別求出不等式組中不等式的解集,利用同大取大,同小取小,大大小小無(wú)解,大小小大取中間,求出不等式組的解集,找出解集中的整數(shù)解即可.
解答:解:(1)原式=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191550493625375/SYS201311011915504936253014_DA/0.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191550493625375/SYS201311011915504936253014_DA/1.png)
+2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191550493625375/SYS201311011915504936253014_DA/2.png)
-3×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191550493625375/SYS201311011915504936253014_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191550493625375/SYS201311011915504936253014_DA/4.png)
;
(2)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191550493625375/SYS201311011915504936253014_DA/5.png)
,
由①去分母得:x-3+6≥2x+2,
解得:x≤1;
由②去括號(hào)得:1-3x+3<8-x,
解得:x>-2,
故不等式組的解集為-2<x≤1,
則不等式組的整數(shù)解為-1,0,1.
點(diǎn)評(píng):此題考查了實(shí)數(shù)的運(yùn)算,以及一元一次不等式組的整數(shù)解,涉及的知識(shí)有:零指數(shù)冪、負(fù)指數(shù)冪,特殊角的三角函數(shù)值,絕對(duì)值的代數(shù)意義,熟練掌握法則是解本題的關(guān)鍵.