题目内容
(1)(x+2y)2-(x+y)(x-y),其中x=-2,y=
(2)[(xy+2)(xy-2)-2(x2y2-2)]÷(xy),其中x=10,y=-
.
| 1 |
| 2 |
(2)[(xy+2)(xy-2)-2(x2y2-2)]÷(xy),其中x=10,y=-
| 1 |
| 25 |
考点:整式的混合运算—化简求值
专题:
分析:(1)先算乘法,再合并同类项,最后代入求出即可;
(2)先算乘法,再合并同类项,算除法,代入求出即可.
(2)先算乘法,再合并同类项,算除法,代入求出即可.
解答:解:(1)(x+2y)2-(x+y)(x-y)
=x2+4xy+4y2-x2+y2
=4xy+5y2,
当x=-2,y=
时,原式=4×(-2)×
+5×(
)2=-
;
(2)[(xy+2)(xy-2)-2(x2y2-2)]÷(xy)
=[x2y2-4-2x2y2+4]÷(xy)
=-x2y2÷xy
=-xy,
当x=10,y=-
时,原式=-10×(-
)=
.
=x2+4xy+4y2-x2+y2
=4xy+5y2,
当x=-2,y=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 11 |
| 4 |
(2)[(xy+2)(xy-2)-2(x2y2-2)]÷(xy)
=[x2y2-4-2x2y2+4]÷(xy)
=-x2y2÷xy
=-xy,
当x=10,y=-
| 1 |
| 25 |
| 1 |
| 25 |
| 2 |
| 5 |
点评:本题考查了整式的混合运算和求值的应用,主要考查学生的计算能力和化简能力,注意运算顺序,难度适中.
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