题目内容
(2002•内江)已知:![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_ST/1.png)
【答案】分析:由已知条件可得:∵
,∴x=a+
+2,x-2=a+
,(x-2)2=(a+
)2即:x2-4x=a2+
-2=(a-
)2,化简原式,并代入求值,由a的取值范围确定式子的值.
解答:解:∵
,
∴x=a+
+2,
x-2=a+
,(x-2)2=(a+
)2
即:x2-4x=a2+
-2=(a-
)2
∴原式=
=(x-2)2-
=(a+
)2-
,
∵0<a<1,∴a-
<0,
∴原式=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/17.png)
=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/18.png)
=a2+2.
点评:此题用了整体代入得数学思想,难度较大.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/5.png)
解答:解:∵
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/6.png)
∴x=a+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/7.png)
x-2=a+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/9.png)
即:x2-4x=a2+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/11.png)
∴原式=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/12.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/13.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/14.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/15.png)
∵0<a<1,∴a-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/16.png)
∴原式=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/17.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291028_DA/18.png)
=a2+2.
点评:此题用了整体代入得数学思想,难度较大.
![](http://thumb2018.1010pic.com/images/loading.gif)
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