题目内容
先化简,再求值:(1)5a2b-[2ab2-2(ab-
| 5 |
| 2 |
| 1 |
| 2 |
(2)-(2xy-y2+
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
分析:(1)(2)两式都应先去括号,再合并同类项,将整式化为最简后再把a、b、x、y的值代入即可.
解答:解:(1)原式=5a2b-2ab2+2ab-5a2b-ab+5ab2
=3ab2+ab,
当a=-6,b=-
时,
原式=3×(-6)×
+(-6)×(-
),
=-
+
,
=-
;
(2)原式=-2xy+y2-
x2-
xy-
y2+2x2,
=
x2-
xy+
y2,
当x=-1,y=-2时,
原式=
-
×(-1)×(-2)+
×4,
=
-
+2,
=-
.
=3ab2+ab,
当a=-6,b=-
| 1 |
| 2 |
原式=3×(-6)×
| 1 |
| 4 |
| 1 |
| 2 |
=-
| 9 |
| 2 |
| 6 |
| 2 |
=-
| 3 |
| 2 |
(2)原式=-2xy+y2-
| 1 |
| 2 |
| 2 |
| 3 |
| 1 |
| 2 |
=
| 3 |
| 2 |
| 8 |
| 3 |
| 1 |
| 2 |
当x=-1,y=-2时,
原式=
| 3 |
| 2 |
| 8 |
| 3 |
| 1 |
| 2 |
=
| 3 |
| 2 |
| 16 |
| 3 |
=-
| 11 |
| 6 |
点评:本题考查了整式的化简.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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