题目内容
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 7×9 |
| 1 |
| 9×11 |
| 1 |
| 11×13 |
| 1 |
| 13×15 |
考点:分数的拆项
专题:
分析:根据题意,由分数的拆项可得,
=
×(
-
);
=
×(
-
);
=
×(
-
);
=
×(
-
);
=
×(
-
);
=
×(
-
);然后再进一步计算即可.
| 1 |
| 3×5 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 5×7 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 7×9 |
| 1 |
| 2 |
| 1 |
| 7 |
| 1 |
| 9 |
| 1 |
| 9×11 |
| 1 |
| 2 |
| 1 |
| 9 |
| 1 |
| 11 |
| 1 |
| 11×13 |
| 1 |
| 2 |
| 1 |
| 11 |
| 1 |
| 13 |
| 1 |
| 13×15 |
| 1 |
| 2 |
| 1 |
| 13 |
| 1 |
| 15 |
解答:
解:
+
+
+
+
+
,
=
×(
-
)+
×(
-
)+
×(
-
)+
×(
-
)+
×(
-
)+
×(
-
),
=
×(
-
+
-
+
-
+
-
+
-
+
-
),
=
×(
-
),
=
×
,
=
.
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 7×9 |
| 1 |
| 9×11 |
| 1 |
| 11×13 |
| 1 |
| 13×15 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 2 |
| 1 |
| 7 |
| 1 |
| 9 |
| 1 |
| 2 |
| 1 |
| 9 |
| 1 |
| 11 |
| 1 |
| 2 |
| 1 |
| 11 |
| 1 |
| 13 |
| 1 |
| 2 |
| 1 |
| 13 |
| 1 |
| 15 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 7 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 11 |
| 1 |
| 11 |
| 1 |
| 13 |
| 1 |
| 13 |
| 1 |
| 15 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 15 |
=
| 1 |
| 2 |
| 4 |
| 15 |
=
| 2 |
| 15 |
点评:本题主要考查分数的拆项,根据分数的拆项公式
=
×(
-
)进行计算即可.
| 1 |
| (2n-1)×(2n+1) |
| 1 |
| 2 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
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