题目内容
|
|
| ||||||||||||
|
|
|
分析:(1)
÷x=
,根据乘法、除法之间的关系,除数等于被除数除以商,由此得解.
(2)
x+4=6,先把方程两边同时减4,再把方程两边同时除以
即可.
(3)
x=16,方程两边同时除以
便得解.
(4)
x+
=
,先把方程两边同时减
,再把方程两边同时除以
.
(5)
÷x=9,根据乘法、除法之间的关系,除数等于被除数除以商,由此得解.
(6)
x=
.方程两边同时除以
即可.
| 4 |
| 5 |
| 2 |
| 15 |
(2)
| 2 |
| 3 |
| 2 |
| 3 |
(3)
| 2 |
| 3 |
| 2 |
| 3 |
(4)
| 1 |
| 3 |
| 1 |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
(5)
| 9 |
| 2 |
(6)
| 1 |
| 2 |
| 8 |
| 9 |
| 1 |
| 2 |
解答:解:(1)
÷x=
,
x=
÷
,
x=
×
,
x=6;
(2)
x+4=6,
x+4-4=6-4,
x÷
=2÷
,
x=3;
(3)
x=16,
x÷
=16÷
,
x=24;
(4)
x+
=
,
x+
-
=
-
,
x÷
=
÷
,
x=1;
(5)
÷x=9,
x=
÷9,
x=
×
,
x=
;
(6)
x=
.
x÷
=
÷
,
x=
.
| 4 |
| 5 |
| 2 |
| 15 |
x=
| 4 |
| 5 |
| 2 |
| 15 |
x=
| 4 |
| 5 |
| 15 |
| 2 |
x=6;
(2)
| 2 |
| 3 |
| 2 |
| 3 |
| 2 |
| 3 |
| 2 |
| 3 |
| 2 |
| 3 |
x=3;
(3)
| 2 |
| 3 |
| 2 |
| 3 |
| 2 |
| 3 |
| 2 |
| 3 |
x=24;
(4)
| 1 |
| 3 |
| 1 |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
x=1;
(5)
| 9 |
| 2 |
x=
| 9 |
| 2 |
x=
| 9 |
| 2 |
| 1 |
| 9 |
x=
| 1 |
| 2 |
(6)
| 1 |
| 2 |
| 8 |
| 9 |
| 1 |
| 2 |
| 1 |
| 2 |
| 8 |
| 9 |
| 1 |
| 2 |
x=
| 16 |
| 9 |
点评:此题考查的目的是理解掌握根据等式的性质解方程的方法步骤.
练习册系列答案
相关题目