题目内容
观察下列等式:
=
-
=
-
…
(1)猜想:
=
-
-
.
(2)直接写出下列各式的结果:
①
+
+
+…+
=
.
②
+
+
+…+
=
.
1 |
2×3 |
1 |
2 |
1 |
3 |
1 |
3×4 |
1 |
3 |
1 |
4 |
…
(1)猜想:
1 |
n(n+1) |
1 |
n |
1 |
n+1 |
1 |
n |
1 |
n+1 |
(2)直接写出下列各式的结果:
①
1 |
1×2 |
1 |
2×3 |
1 |
3×4 |
1 |
2009×2010 |
2009 |
2010 |
2009 |
2010 |
②
1 |
1×2 |
1 |
2×3 |
1 |
3×4 |
1 |
n(n+1) |
n |
n+1 |
n |
n+1 |
分析:(1)先根据题中所给出的列子进行猜想,写出猜想结果即可;
(2)根据(1)中的猜想计算出结果.
(2)根据(1)中的猜想计算出结果.
解答:解:(1)∵
=
-
;
=
-
;
∴
=
-
.
故答案为:
-
;
(2)①原式=1-
+
-
+
-
+…+
-
=1-
=
;
故答案为:
;
②原式═1-
+
-
+
-
+…+
-
=1-
=
.
故答案为:
.
1 |
2×3 |
1 |
2 |
1 |
3 |
1 |
3×4 |
1 |
3 |
1 |
4 |
∴
1 |
n(n+1) |
1 |
n |
1 |
n+1 |
故答案为:
1 |
n |
1 |
n+1 |
(2)①原式=1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
2009 |
1 |
2010 |
=1-
1 |
2010 |
=
2009 |
2010 |
故答案为:
2009 |
2010 |
②原式═1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
n |
1 |
n+1 |
=1-
1 |
n+1 |
=
n |
n+1 |
故答案为:
n |
n+1 |
点评:本题考查的是分式的加减,根据题意找出规律是解答此题的关键.
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