题目内容
(2003•安徽)已知:x=-1,y=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232335035682924/SYS201310212323350356829010_ST/0.png)
【答案】分析:因为x2+y2-xy中,没有同类项,所以直接代入求值即可.
解答:解:当x=-1,y=
时
x2+y2-xy=(-1)2+(
)2-(-1)×![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232335035682924/SYS201310212323350356829010_DA/2.png)
=3+
.
点评:本题主要考查了整式的代入求值问题,属于基础题型.解题时要正确处理符号问题.
解答:解:当x=-1,y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232335035682924/SYS201310212323350356829010_DA/0.png)
x2+y2-xy=(-1)2+(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232335035682924/SYS201310212323350356829010_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232335035682924/SYS201310212323350356829010_DA/2.png)
=3+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232335035682924/SYS201310212323350356829010_DA/3.png)
点评:本题主要考查了整式的代入求值问题,属于基础题型.解题时要正确处理符号问题.
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目