题目内容
先化简,再求值:(1)2(2a2-2a-5)-(4a2-3a-6),其中a=-3;
(2)-x2+[3xy2-
(4x2-8xy2)+x2]-7xy2,其中x=-
,y=-2011.
(2)-x2+[3xy2-
| 1 |
| 2 |
| 3 |
| 2 |
分析:(1)本题应对整式合并同类项,将整式化为最简式,再将a的值代入即可.
(2)本题应对整式去括号,合并同类项,将整式化为最简式,然后把x,y的值代入即可.
(2)本题应对整式去括号,合并同类项,将整式化为最简式,然后把x,y的值代入即可.
解答:解:(1)2(2a2-2a-5)-(4a2-3a-6)
=4a2-4a-10-4a2+3a+6
=-a-4,
当a=-3时-a-4=-(-3)-4=-1;
(2)-x2+[3xy2-
(4x2-8xy2)+x2]-7xy2
=-x2+[3xy2-2x2-4xy2)+x2]-7xy2
=-x2+[7xy2-x2]-7xy2
=-x2+7xy2-x2-7xy2=-2x2
当x=-
,y=-2011时,
-2x2=-2×(-
)2=-
.
=4a2-4a-10-4a2+3a+6
=-a-4,
当a=-3时-a-4=-(-3)-4=-1;
(2)-x2+[3xy2-
| 1 |
| 2 |
=-x2+[3xy2-2x2-4xy2)+x2]-7xy2
=-x2+[7xy2-x2]-7xy2
=-x2+7xy2-x2-7xy2=-2x2
当x=-
| 3 |
| 2 |
-2x2=-2×(-
| 3 |
| 2 |
| 9 |
| 2 |
点评:本题考查了整式的化简求值.整式的加减运算实际上就是去括号、合并同类项,这是各地中考的常考点.
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