题目内容
解方程.
(1)x÷
|
(2)
|
(3)
|
(4)
|
分析:(1)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x,得
×x=
,两边同乘
即可;
(3)根据等式的性质,两边同减去
,再同乘4即可;
(4)根据等式的性质,两边同加上3x,得
+3x=
,两边同减去
,再同除以3即可.
| 6 |
| 25 |
(2)根据等式的性质,两边同乘x,得
| 14 |
| 27 |
| 7 |
| 18 |
| 27 |
| 14 |
(3)根据等式的性质,两边同减去
| 3 |
| 4 |
(4)根据等式的性质,两边同加上3x,得
| 3 |
| 14 |
| 6 |
| 7 |
| 3 |
| 14 |
解答:解:(1)x÷
=
,
x÷
×
=
×
,
x=
;
(2)
÷x=
,
÷x×x=
×x,
×x=
,
×x×
=
×
,
x=
;
(3)
x+
=
,
x+
-
=
-
,
x=
,
x×4=
×4,
x=
;
(4)
-3x=
,
-3x+3x=
+3x,
+3x=
,
+3x-
=
-
,
3x=
,
3x÷3=
÷3,
x=
.
| 6 |
| 25 |
| 5 |
| 12 |
x÷
| 6 |
| 25 |
| 6 |
| 25 |
| 5 |
| 12 |
| 6 |
| 25 |
x=
| 1 |
| 10 |
(2)
| 7 |
| 18 |
| 14 |
| 27 |
| 7 |
| 18 |
| 14 |
| 27 |
| 14 |
| 27 |
| 7 |
| 18 |
| 14 |
| 27 |
| 27 |
| 14 |
| 7 |
| 18 |
| 27 |
| 14 |
x=
| 3 |
| 4 |
(3)
| 1 |
| 4 |
| 3 |
| 4 |
| 5 |
| 6 |
| 1 |
| 4 |
| 3 |
| 4 |
| 3 |
| 4 |
| 5 |
| 6 |
| 3 |
| 4 |
| 1 |
| 4 |
| 1 |
| 12 |
| 1 |
| 4 |
| 1 |
| 12 |
x=
| 1 |
| 3 |
(4)
| 6 |
| 7 |
| 3 |
| 14 |
| 6 |
| 7 |
| 3 |
| 14 |
| 3 |
| 14 |
| 6 |
| 7 |
| 3 |
| 14 |
| 3 |
| 14 |
| 6 |
| 7 |
| 3 |
| 14 |
3x=
| 9 |
| 14 |
3x÷3=
| 9 |
| 14 |
x=
| 3 |
| 14 |
点评:此题考查了根据等式的性质解方程,即等式两边同加上、同减去、同乘上或同除以一个不为0的数,等式仍相等.同时注意“=”上下要对齐.
练习册系列答案
相关题目