题目内容
先观察1-
=
×
,1-
=
×
,1-
=
×
(1)按上述规律填空:1-
=
×
;1-
=
×
.
(2)计算:(1-
)•(1-
)•(1-
)•…•(1-
).
| 1 |
| 22 |
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 32 |
| 2 |
| 3 |
| 4 |
| 3 |
| 1 |
| 42 |
| 3 |
| 4 |
| 5 |
| 4 |
(1)按上述规律填空:1-
| 1 |
| 1002 |
| 99 |
| 100 |
| 99 |
| 100 |
| 101 |
| 100 |
| 101 |
| 100 |
| 1 |
| 20102 |
| 2009 |
| 2010 |
| 2009 |
| 2010 |
| 2011 |
| 2010 |
| 2011 |
| 2010 |
(2)计算:(1-
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| 20102 |
分析:(1)观察已知等式可知,等式右边为两个分数的积,其分母相等且与等式左边分母的底数相等,分子一个比分母小1,一个比分母大1,由此填空;
(2)根据(1)发现的规律,将每个括号部分分解为两个分数的积,再寻找约分规律.
(2)根据(1)发现的规律,将每个括号部分分解为两个分数的积,再寻找约分规律.
解答:解:(1)依题意,得1-
=
×
,1-
=
×
,
故答案为:
,
,
,
;
(2)原式=
×
×
×
×
×
×…×
×
=
×
=
.
| 1 |
| 1002 |
| 99 |
| 100 |
| 101 |
| 100 |
| 1 |
| 20102 |
| 2009 |
| 2010 |
| 2011 |
| 2010 |
故答案为:
| 99 |
| 100 |
| 101 |
| 100 |
| 2009 |
| 2010 |
| 2011 |
| 2010 |
(2)原式=
| 1 |
| 2 |
| 3 |
| 2 |
| 2 |
| 3 |
| 4 |
| 3 |
| 3 |
| 4 |
| 5 |
| 4 |
| 2009 |
| 2010 |
| 2011 |
| 2010 |
=
| 1 |
| 2 |
| 2011 |
| 2010 |
=
| 2011 |
| 4020 |
点评:本题考查的是有理数的运算能力.关键是根据已知等式,由特殊到一般,得出分数的拆分规律和约分规律.
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