题目内容
先化简,再求值:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_ST/0.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_ST/1.png)
【答案】分析:先把分式的分子和分母因式分解得到原式=
,约分后得原式=
,由a=
-
,b=
+
,先计算出a+b=
-
+
+
=2
,ab=(
-
)(
+
)=
=3-11=-8,然后整体代入a+b与ab的值即可得到原分式的值.
解答:解:原式=![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/16.png)
=
,
当a=
-
,b=
+
时,a+b=
-
+
+
=2
,ab=(
-
)(
+
)=
=3-11=-8,
所以原式=-
=-
.
点评:本题考查了分式的化简求值:先把各分式的分子或分母因式分解,再进行分式的乘除运算,然后进行分式的加减运算得到最简分式或整式,再把满足条件的字母的值代入计算得到对应的分式的值;有括号先算括号.也考查了二次根式的计算.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/0.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/1.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/2.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/3.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/4.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/5.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/6.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/7.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/8.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/9.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/10.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/11.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/12.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/13.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/14.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/15.png)
解答:解:原式=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/17.png)
当a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/18.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/19.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/20.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/21.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/22.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/23.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/24.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/25.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/26.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/27.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/28.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/29.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/30.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/31.png)
所以原式=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/32.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164845022065751/SYS201310221648450220657016_DA/33.png)
点评:本题考查了分式的化简求值:先把各分式的分子或分母因式分解,再进行分式的乘除运算,然后进行分式的加减运算得到最简分式或整式,再把满足条件的字母的值代入计算得到对应的分式的值;有括号先算括号.也考查了二次根式的计算.
![](http://thumb.zyjl.cn/images/loading.gif)
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