题目内容
观察下列各式:| 1 |
| 6 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 12 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 20 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 20 |
| 1 |
| 4 |
| 1 |
| 5 |
(1)由此可以推断
| 1 |
| 30 |
(2)请用上面的规律解方程:
| 1 |
| (x-1)(x-2) |
| 1 |
| (x-2)(x-3) |
| 1 |
| (x-3)(x-4) |
| 3 |
| 4 |
分析:(1)将30分解成为5×6,从而得出
=
-
,
(2)由上面的规律可得
=
-
,
=
-
,
=
-
,再解方程即可
| 1 |
| 30 |
| 1 |
| 5 |
| 1 |
| 6 |
(2)由上面的规律可得
| 1 |
| (x-1)(x-2) |
| 1 |
| x-2 |
| 1 |
| x-1 |
| 1 |
| (x-2)(x-3) |
| 1 |
| x-3 |
| 1 |
| x-2 |
| 1 |
| (x-3)(x-4) |
| 1 |
| x-4 |
| 1 |
| x-3 |
解答:解:(1)
=
-
;
(2)∵
=
-
,
=
-
,
=
-
,
∴原方程可化为
,
方程两边乘以(x-1)(x-4),得
即x2-5x=0
x1=5,x2=0,
检验,把x=5代入(x-1)(x-4)=4≠0,
把x=0代入(x-1)(x-4)=4≠0,
∴x1=5,x2=0,都是方程的解.
| 1 |
| 30 |
| 1 |
| 5 |
| 1 |
| 6 |
(2)∵
| 1 |
| (x-1)(x-2) |
| 1 |
| x-2 |
| 1 |
| x-1 |
| 1 |
| (x-2)(x-3) |
| 1 |
| x-3 |
| 1 |
| x-2 |
| 1 |
| (x-3)(x-4) |
| 1 |
| x-4 |
| 1 |
| x-3 |
∴原方程可化为
|
方程两边乘以(x-1)(x-4),得
即x2-5x=0
x1=5,x2=0,
检验,把x=5代入(x-1)(x-4)=4≠0,
把x=0代入(x-1)(x-4)=4≠0,
∴x1=5,x2=0,都是方程的解.
点评:本题考查了分式方程的解法,在第(2)中将
化为
-
是解题的关键.
| 1 |
| (x-1)(x-2) |
| 1 |
| x-2 |
| 1 |
| x-1 |
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