题目内容
计算(1-
-
-…-
)(
+
+…+
)-(1-
-
-…-
)(
+
+…+
)的结果应该是( )
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2003 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2004 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2004 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2003 |
A、
| ||
B、
| ||
C、
| ||
D、
|
分析:此题需要仔细观察式子的特点,寻找简便方法,切不可盲目计算.
解答:解:设(1-
-
-…-
)=x;(
+
+…+
)=y
则原式=x(y+
)-(x-
)y
=xy+
-xy+
=
又∵x+y=(1-
-
-…-
)+(
+
+…+
)=1
∴原式=
故选A.
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2003 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2003 |
则原式=x(y+
| 1 |
| 2004 |
| 1 |
| 2004 |
=xy+
| x |
| 2004 |
| y |
| 2004 |
=
| x+y |
| 2004 |
又∵x+y=(1-
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2003 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2003 |
∴原式=
| 1 |
| 2004 |
故选A.
点评:此题考查了有理数的混合运算以及换元的思想.
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