题目内容
若a是方程x2-5x+1=0的一个根,求a2+![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202408604388999/SYS201311032024086043889023_ST/0.png)
【答案】分析:把a代入原方程,得到关于a的一元二次方程,a2-5a+1=0,代入直接求值即可.
解答:解:依题意得,
a2-5a+1=0,则a≠0,方程两边同时除以a,得a-5+
=0,
∴a+
=5,两边同时平方,得:
(a+
)2=25,a2+
+2=25,
∴a2+
=23.
点评:代数式中的字母表示的数没有明确告知,而是隐含在题设中,首先应从题设中获取代数式a2+
的值,然后利用“整体代入法”求代数式的值.
解答:解:依题意得,
a2-5a+1=0,则a≠0,方程两边同时除以a,得a-5+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202408604388999/SYS201311032024086043889023_DA/0.png)
∴a+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202408604388999/SYS201311032024086043889023_DA/1.png)
(a+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202408604388999/SYS201311032024086043889023_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202408604388999/SYS201311032024086043889023_DA/3.png)
∴a2+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202408604388999/SYS201311032024086043889023_DA/4.png)
点评:代数式中的字母表示的数没有明确告知,而是隐含在题设中,首先应从题设中获取代数式a2+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202408604388999/SYS201311032024086043889023_DA/5.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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