题目内容
如果A=2x2+3xy-2x-1,B=-x2+xy-1,且3A+6B的值与x的取值无关,求1 |
1×2 |
1 |
2×3 |
1 |
3×4 |
1 |
4×5 |
1 |
5×6 |
1 |
6×7 |
1 |
7×8 |
1 |
8×9 |
1 |
y |
分析:把A、B代入3A+6B,由3A+6B的值与x的取值无关可求出y的值;把y代入代数式进行计算即可.注意利用
=
-
将式子化简.
1 |
n(n+1) |
1 |
n |
1 |
n+1 |
解答:解:3A+6B=3(2x2+3xy-2x-1)+6(-x2+xy-1)
=6x2+9xy-6x-3-6x2+6xy-6
=15xy-6x-9
=(15y-6)x-9
∵3A+6B的值与x的取值无关,
∴15y=6,即y=
.
∴原式=1-
+
-
+…+
-
-
=1-
-
=
=-
.
=6x2+9xy-6x-3-6x2+6xy-6
=15xy-6x-9
=(15y-6)x-9
∵3A+6B的值与x的取值无关,
∴15y=6,即y=
2 |
5 |
∴原式=1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
8 |
1 |
9 |
1 |
y |
=1-
1 |
9 |
5 |
2 |
=
18-2-45 |
18 |
=-
29 |
18 |
点评:此题考查了整式的加减运算.注意①与某字母的值无关,则式子中不含该字母;②
=
-
.
1 |
n(n+1) |
1 |
n |
1 |
n+1 |
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