题目内容
先化简,再求值:[(x+y)(x-2y)-(x+2y)2]÷| 1 |
| 2 |
| 1 |
| 4 |
分析:本题的关键是化简,然后把给定的值代入求值.
解答:解:[(x+y)(x-2y)-(x+2y)2]÷
y
=[x2-xy-2y2-(x2+4xy+4y2)]÷
y
=[x2-xy-2y2-x2-4xy-4y2]÷
y
=(-5xy-6y2)÷
y
=-10x-12y
把x=-1,y=
代入,原式=-10×(-1)-12×
=10-3=7.
| 1 |
| 2 |
=[x2-xy-2y2-(x2+4xy+4y2)]÷
| 1 |
| 2 |
=[x2-xy-2y2-x2-4xy-4y2]÷
| 1 |
| 2 |
=(-5xy-6y2)÷
| 1 |
| 2 |
=-10x-12y
把x=-1,y=
| 1 |
| 4 |
| 1 |
| 4 |
点评:考查了整式的混合运算,主要考查了完全平方公式、整式的乘法、除法、合并同类项的知识点.注意运算顺序,符号的处理.
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