题目内容
计算:(1-
)(1-
)(1-
)…(1-
)(1-
)(1-
).
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| 982 |
| 1 |
| 992 |
| 1 |
| 1002 |
考点:有理数的混合运算
专题:计算题
分析:原式各括号中利用平方差公式化简,整理后约分即可得到结果.
解答:解:原式=(1+
)(1-
)(1+
)(1-
)…(1+
)(1-
)(1+
)(1-
)(1+
)(1-
)
=
×
×
×
××
×
×
×
×
×
=
×
=
.
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 98 |
| 1 |
| 98 |
| 1 |
| 99 |
| 1 |
| 99 |
| 1 |
| 100 |
| 1 |
| 100 |
=
| 1 |
| 2 |
| 3 |
| 2 |
| 2 |
| 3 |
| 4 |
| 3 |
| 97 |
| 98 |
| 99 |
| 98 |
| 98 |
| 99 |
| 100 |
| 99 |
| 99 |
| 100 |
| 101 |
| 100 |
=
| 1 |
| 2 |
| 101 |
| 100 |
=
| 101 |
| 200 |
点评:此题考查了有理数的混合运算,熟练掌握运算法则是解本题的关键.
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