题目内容
解方程.
X-
|
X÷
|
|
分析:(1)原式变为(1-
)X=
,即
X=
,根据等式的性质,两边同乘
即可;
(2)根据等式的性质,两边同乘
即可;
(3)根据等式的性质,两边同乘
,再同乘
即可.
| 1 |
| 4 |
| 3 |
| 8 |
| 3 |
| 4 |
| 3 |
| 8 |
| 4 |
| 3 |
(2)根据等式的性质,两边同乘
| 4 |
| 5 |
(3)根据等式的性质,两边同乘
| 1 |
| 4 |
| 3 |
| 2 |
解答:解:(1)X-
X=
,
(1-
)X=
,
X=
,
X×
=
×
,
X=
;
(2)X÷
=
,
X÷
×
=
×
,
X=
;
(3)
X÷
=12,
X÷
×
=12×
,
X=3,
X×
=3×
,
X=
.
| 1 |
| 4 |
| 3 |
| 8 |
(1-
| 1 |
| 4 |
| 3 |
| 8 |
| 3 |
| 4 |
| 3 |
| 8 |
| 3 |
| 4 |
| 4 |
| 3 |
| 3 |
| 8 |
| 4 |
| 3 |
X=
| 1 |
| 2 |
(2)X÷
| 4 |
| 5 |
| 15 |
| 28 |
X÷
| 4 |
| 5 |
| 4 |
| 5 |
| 15 |
| 28 |
| 4 |
| 5 |
X=
| 3 |
| 7 |
(3)
| 2 |
| 3 |
| 1 |
| 4 |
| 2 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 4 |
| 2 |
| 3 |
| 2 |
| 3 |
| 3 |
| 2 |
| 3 |
| 2 |
X=
| 9 |
| 2 |
点评:在解方程时应根据等式的性质,即等式两边同加上、同减去、同乘上或同除以某一个数(0除外),等式的两边仍相等,同时注意“=”上下要对齐.
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