题目内容
先化简,再求值:
(1)(3m+1)(2m-3)-(6m-5)(m-4),其中m=-
;
(2)[xy-2x(y-
x)]•3x+3y(x2-y2),其中x=-1,y=
.
(1)(3m+1)(2m-3)-(6m-5)(m-4),其中m=-
| 1 |
| 2 |
(2)[xy-2x(y-
| 1 |
| 3 |
| 1 |
| 3 |
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,最后代入求出即可.
(2)先算乘法,再合并同类项,最后代入求出即可.
解答:解:(1)(3m+1)(2m-3)-(6m-5)(m-4)
=6m2-9m+2m-3-6m2+24m+5m-20
=22m-23,
当m=-
时,
原式=22×(-
)-23=-34.
(2)[xy-2x(y-
x)]•3x+3y(x2-y2)
=[xy-2xy+
x2]•3x+3y(x2-y2)
=-3x2y+2x3+3x2y-3y3
=2x3-3y3
当x=-1,y=
时,
原式=2×(-1)3-3×(
)3=-2
.
=6m2-9m+2m-3-6m2+24m+5m-20
=22m-23,
当m=-
| 1 |
| 2 |
原式=22×(-
| 1 |
| 2 |
(2)[xy-2x(y-
| 1 |
| 3 |
=[xy-2xy+
| 2 |
| 3 |
=-3x2y+2x3+3x2y-3y3
=2x3-3y3
当x=-1,y=
| 1 |
| 3 |
原式=2×(-1)3-3×(
| 1 |
| 3 |
| 1 |
| 9 |
点评:本题考查了整式的混合运算和求值的应用,主要考查学生的化简和计算能力.
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