题目内容
观察:
=
=
-
,
=
=
-
,
=
=
-
,
=
=
-
,
=
=
-
,…
(1)猜想:请你猜想出表示(1)中的特点的一般规律,用含x(x表示整数)的等式表示出来
=
-
(x表示整数)
=
-
(x表示整数).
(2)验证:
(3)运用:请利用上述规律,解方程
+
+
+
+
=
.
| 1 |
| 2 |
| 1 |
| 1×2 |
| 1 |
| 1 |
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 2×3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 12 |
| 1 |
| 3×4 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 20 |
| 1 |
| 4×5 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 30 |
| 1 |
| 5×6 |
| 1 |
| 5 |
| 1 |
| 6 |
(1)猜想:请你猜想出表示(1)中的特点的一般规律,用含x(x表示整数)的等式表示出来
| 1 |
| x(x+1) |
| 1 |
| x |
| 1 |
| x+1 |
| 1 |
| x(x+1) |
| 1 |
| x |
| 1 |
| x+1 |
(2)验证:
(3)运用:请利用上述规律,解方程
| 1 |
| (x-4)(x-3) |
| 1 |
| (x-3)(x-2) |
| 1 |
| (x-2)(x-1) |
| 1 |
| (x-1)x |
| 1 |
| x(x+1) |
| 1 |
| x+1 |
分析:(1)根据题意观察,即可得到规律:
=
-
(x表示整数);
(2)利用分式的加减运算,即可验证猜想的准确性;
(3)利用规律,将原方程化为:
-
=
,解此分式方程即可求得答案.
| 1 |
| x(x+1) |
| 1 |
| x |
| 1 |
| x+1 |
(2)利用分式的加减运算,即可验证猜想的准确性;
(3)利用规律,将原方程化为:
| 1 |
| x-4 |
| 1 |
| x+1 |
| 1 |
| x+1 |
解答:解:观察::
=
=
-
,
=
=
-
,
=
=
-
,
=
=
-
,
=
=
-
,…
(1)猜想:请你猜想出表示(1)中的特点的一般规律,用含x(x表示整数)的等式表示出来
=
-
(x表示整数);
(2)验证:右边=
-
=
-
=
=
=左边,
故猜想正确;
(3)原方程可变形为:
-
+
-
+
-
+
-
+
-
=
,
则
-
=
,
解得x=9.
经检验:方程的根是x=9;
故原方程的根为:x=9.
故答案为:
=
-
(x表示整数).
| 1 |
| 2 |
| 1 |
| 1×2 |
| 1 |
| 1 |
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 2×3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 12 |
| 1 |
| 3×4 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 20 |
| 1 |
| 4×5 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 30 |
| 1 |
| 5×6 |
| 1 |
| 5 |
| 1 |
| 6 |
(1)猜想:请你猜想出表示(1)中的特点的一般规律,用含x(x表示整数)的等式表示出来
| 1 |
| x(x+1) |
| 1 |
| x |
| 1 |
| x+1 |
(2)验证:右边=
| 1 |
| x |
| 1 |
| x+1 |
| x+1 |
| x(x+1) |
| x |
| x(x+1) |
| x+1-x |
| x(x+1) |
| 1 |
| x(x+1) |
故猜想正确;
(3)原方程可变形为:
| 1 |
| x-4 |
| 1 |
| x-3 |
| 1 |
| x-3 |
| 1 |
| x-2 |
| 1 |
| x-2 |
| 1 |
| x-1 |
| 1 |
| x-1 |
| 1 |
| x |
| 1 |
| x |
| 1 |
| x+1 |
| 1 |
| x+1 |
则
| 1 |
| x-4 |
| 1 |
| x+1 |
| 1 |
| x+1 |
解得x=9.
经检验:方程的根是x=9;
故原方程的根为:x=9.
故答案为:
| 1 |
| x(x+1) |
| 1 |
| x |
| 1 |
| x+1 |
点评:此题考查了分式方程与分式的运算等知识.注意找到规律:
=
-
是解此题的关键.
| 1 |
| x(x+1) |
| 1 |
| x |
| 1 |
| x+1 |
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