题目内容
先观察下列等式,然后用你发现的规律解答下面问题
=1-
=
-
=
-
(1)填空
+
+
+…+
=
;
(2)
+
+
+…+
;
(3)如果将问题改为如下形式,你还会计算吗?
+
+
;
(4)解方程
+
+
+…+
=503.
| 1 |
| 1×2 |
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3×4 |
| 1 |
| 3 |
| 1 |
| 4 |
(1)填空
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 9×10 |
| 9 |
| 10 |
| 9 |
| 10 |
(2)
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| (n-1)n |
(3)如果将问题改为如下形式,你还会计算吗?
| 1 |
| 1×5 |
| 1 |
| 5×9 |
| 1 |
| 9×13 |
(4)解方程
| x |
| 1×5 |
| x |
| 5×9 |
| x |
| 9×13 |
| x |
| 2009×2013 |
分析:(1)类比题目中的拆项方法,类比得出答案即可;
(2)利用(1)的结论进一步推广为一般形式得出结果;
(3)分母是相差4的两个自然数的乘积,类比拆成以两个自然数为分母,分子为1的两个自然数差的
即可,得出结论;
(4)利用(3)的方法转化为一元一次方程求出解即可.
(2)利用(1)的结论进一步推广为一般形式得出结果;
(3)分母是相差4的两个自然数的乘积,类比拆成以两个自然数为分母,分子为1的两个自然数差的
| 1 |
| 4 |
(4)利用(3)的方法转化为一元一次方程求出解即可.
解答:解:(1)
+
+
+…+
=1-
+
-
+
-
+…+
-
=1-
=
;
(2)
+
+
+…+
=1-
+
-
+
-
+…+
-
=1-
=
;
(3)
+
+
=
×(1-
+
-
+
-
)
=
×
=
;
(4)
+
+
+…+
=503
×(1-
+
-
+
-
+…+
-
)x=503
×
x=503
x=2013.
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 9×10 |
=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 9 |
| 1 |
| 10 |
=1-
| 1 |
| 10 |
=
| 9 |
| 10 |
(2)
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| (n-1)n |
=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n-1 |
| 1 |
| n |
=1-
| 1 |
| n |
=
| n-1 |
| n |
(3)
| 1 |
| 1×5 |
| 1 |
| 5×9 |
| 1 |
| 9×13 |
=
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 13 |
=
| 1 |
| 4 |
| 12 |
| 13 |
=
| 3 |
| 13 |
(4)
| x |
| 1×5 |
| x |
| 5×9 |
| x |
| 9×13 |
| x |
| 2009×2013 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 13 |
| 1 |
| 2009 |
| 1 |
| 2013 |
| 1 |
| 4 |
| 2012 |
| 2013 |
x=2013.
点评:此题考查算式的规律,注意分数的分母、分子的特点,灵活进行拆项,进一步利用规律解决问题.
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