题目内容
观察下列各式,回答问题
1-
=
×
,1-
=
×
,1-
=
×
….
按上述规律填空:
(1)1-
=
×
,1-
=
×
.
(2)计算:(1-
)×(1-
)…×(1-
)×(1-
).
1-
| 1 |
| 22 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 32 |
| 2 |
| 3 |
| 4 |
| 3 |
| 1 |
| 42 |
| 3 |
| 4 |
| 5 |
| 4 |
按上述规律填空:
(1)1-
| 1 |
| 1002 |
| 99 |
| 100 |
| 99 |
| 100 |
| 101 |
| 100 |
| 101 |
| 100 |
| 1 |
| 20052 |
| 2004 |
| 2005 |
| 2004 |
| 2005 |
| 2006 |
| 2005 |
| 2006 |
| 2005 |
(2)计算:(1-
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 20042 |
| 1 |
| 20052 |
分析:首先可以看出等号的左边是1减去几的平方分之一,计算的结果是1减去几分之一乘1加上几分之一,由此规律直接得出答案即可.
解答:解:(1)1-
=
×
,1-
=
×
.
(2)(1-
)×(1-
)…×(1-
)×(1-
)
=
×
×
×
×
×
×
×
=
×
.
=
.
故答案为:
,
;
,
.
| 1 |
| 1002 |
| 99 |
| 100 |
| 101 |
| 100 |
| 1 |
| 20052 |
| 2004 |
| 2005 |
| 2006 |
| 2005 |
(2)(1-
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 20042 |
| 1 |
| 20052 |
=
| 1 |
| 2 |
| 3 |
| 2 |
| 2 |
| 3 |
| 4 |
| 3 |
| 2003 |
| 2004 |
| 2005 |
| 2004 |
| 2004 |
| 2005 |
| 2006 |
| 2005 |
=
| 1 |
| 2 |
| 2006 |
| 2005 |
=
| 1003 |
| 2005 |
故答案为:
| 99 |
| 100 |
| 101 |
| 100 |
| 2004 |
| 2005 |
| 2006 |
| 2005 |
点评:此题考查有理数的混合运算,从最简单的情形入手,找出规律,利用规律简化计算的方法.
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