题目内容
计算:(1+
)•(1+
)•(1+
)•(1+
)•
.
| 1 |
| 2 |
| 1 |
| 22 |
| 1 |
| 24 |
| 1 |
| 28 |
| 1 |
| 215 |
考点:平方差公式
专题:
分析:先根据平方差公式变形,再根据平方差公式依次进行计算即可.
解答:解:原式=
•(1-
)•(1+
)•(1+
)•(1+
)•(1+
)•
═
•(1-
)•(1+
)•(1+
)•(1+
)•
=
•(1-
)•(1+
)•(1+
)•
═
•(1-
)•(1+
)•
═
•(1-
)•
=
-
=
.
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 22 |
| 1 |
| 24 |
| 1 |
| 28 |
| 1 |
| 215 |
═
| 1 |
| 2 |
| 1 |
| 22 |
| 1 |
| 22 |
| 1 |
| 24 |
| 1 |
| 28 |
| 1 |
| 215 |
=
| 1 |
| 2 |
| 1 |
| 24 |
| 1 |
| 24 |
| 1 |
| 28 |
| 1 |
| 215 |
═
| 1 |
| 2 |
| 1 |
| 28 |
| 1 |
| 28 |
| 1 |
| 215 |
═
| 1 |
| 2 |
| 1 |
| 216 |
| 1 |
| 215 |
=
| 1 |
| 216 |
| 1 |
| 232 |
=
| 216-1 |
| 232 |
点评:本题考查了平方差公式的应用,主要考查学生能否灵活运用平方差公式进行计算,难度适中.
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