题目内容
先化简,再求值:
(1)2(x-y)-3(
x-2y-1),其中x=2011,y=-
.
(2)-3(x2-xy)+2(x2+xy-1)+1 其中x=
,y=-2.
(1)2(x-y)-3(
| 1 |
| 3 |
| 1 |
| 4 |
(2)-3(x2-xy)+2(x2+xy-1)+1 其中x=
| 1 |
| 2 |
分析:(1)先去括号,再合并同类项,然后把x、y的值代入进行计算即可得解;
(2)先去括号,再合并同类项,然后把x、y的值代入进行计算即可得解.
(2)先去括号,再合并同类项,然后把x、y的值代入进行计算即可得解.
解答:解:(1)2(x-y)-3(
x-2y-1)
=2x-2y-x+6y+3
=(2-1)x+(-2+6)y+3
=x+4y+3,
当x=2001,y=-
时,原式=2011+4×(-
)+3=2011-1+3=2003;
(2)-3(x2-xy)+2(x2+xy-1)+1
=-3x2+3xy+2x2+2xy-2+1
=(-3+2)x2+(3+2)xy+(-2+1)
=-x2+5xy-1,
当x=
,y=-2时,原式=-(
)2+5×
×(-2)-1=-
-5-1=-
.
| 1 |
| 3 |
=2x-2y-x+6y+3
=(2-1)x+(-2+6)y+3
=x+4y+3,
当x=2001,y=-
| 1 |
| 4 |
| 1 |
| 4 |
(2)-3(x2-xy)+2(x2+xy-1)+1
=-3x2+3xy+2x2+2xy-2+1
=(-3+2)x2+(3+2)xy+(-2+1)
=-x2+5xy-1,
当x=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
| 25 |
| 4 |
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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