题目内容
(2006•南通)(1)计算![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_ST/0.png)
(2)解不等式组
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_ST/1.png)
【答案】分析:(1)根据数的开方,分母有理化及非0数的幂的运算法则来计算;
(2)求出两个不等式的解集,再求其公共解.
解答:解:(1)原式=3
-
-
+1
=3
-
-
+1
=
+1;
(2)![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_DA/7.png)
由①得x+1>3-x,即x>1;
由②得4x+16<3x+18,即x<2;
不等式组的解集为1<x<2.
点评:实数运算要先算乘方,再算乘除,后算加减.求不等式的公共解,要遵循以下原则:同大取较大,同小取较小,小大大小中间找,大大小小解不了.
(2)求出两个不等式的解集,再求其公共解.
解答:解:(1)原式=3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_DA/2.png)
=3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_DA/5.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_DA/6.png)
(2)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231852314169923/SYS201310212318523141699028_DA/7.png)
由①得x+1>3-x,即x>1;
由②得4x+16<3x+18,即x<2;
不等式组的解集为1<x<2.
点评:实数运算要先算乘方,再算乘除,后算加减.求不等式的公共解,要遵循以下原则:同大取较大,同小取较小,小大大小中间找,大大小小解不了.
![](http://thumb2018.1010pic.com/images/loading.gif)
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