题目内容
先阅读下列材料,再解答后面的问题.∵
| 1 |
| 1×3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3×5 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 5×7 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 17×19 |
| 1 |
| 2 |
| 1 |
| 17 |
| 1 |
| 19 |
∴
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 17×19 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 2 |
| 1 |
| 17 |
| 1 |
| 19 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 17 |
| 1 |
| 19 |
=
| 1 |
| 2 |
| 1 |
| 19 |
| 9 |
| 19 |
(1)在和式
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 5×7 |
(2)计算
| 1 |
| (x-1)x |
| 1 |
| x(x+1) |
| 1 |
| (x+1)(x+2) |
| 1 |
| (x+2)(x+3) |
分析:(1)由已知中所给的条件找出规律解答;(2)把每一个分式先分列为两个分式,找抵消规律,再计算.
解答:解:(1)在和式
+
+
中,
第五项为:
,
第n项为:
;
(2)
+
+
+
=
-
+
-
+
-
+
-
=
-
=
=
.
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 5×7 |
第五项为:
| 1 |
| 9×11 |
第n项为:
| 1 |
| (2n-1)(2n+1) |
(2)
| 1 |
| (x-1)x |
| 1 |
| x(x+1) |
| 1 |
| (x+1)(x+2) |
| 1 |
| (x+2)(x+3) |
=
| 1 |
| x-1 |
| 1 |
| x |
| 1 |
| x |
| 1 |
| x+1 |
| 1 |
| x+1 |
| 1 |
| x+2 |
| 1 |
| x+2 |
| 1 |
| x+3 |
=
| 1 |
| x-1 |
| 1 |
| x+3 |
=
| x+3-x+1 |
| (x-1)(x+3) |
=
| 4 |
| (x-1)(x+3) |
点评:解答此类题目关键是找出规律再解答;在计算分式时若分母有规律可循是可将其分开以简化计算.
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