题目内容
求未知数x.
(1)1-
=
(2)42:
=x:
.
(1)1-
| x |
| 3 |
| 2 |
| 3 |
(2)42:
| 1 |
| 3 |
| 5 |
| 7 |
考点:方程的解和解方程,解比例
专题:简易方程
分析:①依据等式的性质,方程两边同时加
,再同时减
,再同时除以
解答即可.
②先根据比例的基本性质,把原式转化为
x=42×
,再根据等式的性质,在方程两边同时除以
解答.
| x |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
②先根据比例的基本性质,把原式转化为
| 1 |
| 3 |
| 5 |
| 7 |
| 1 |
| 3 |
解答:
解:①1-
=
1-
+
=
+
1-
=
+
-
=
=
÷
=
÷
x=1
②42:
=x:
x=42×
x=30
x÷
=30÷
x=90
| x |
| 3 |
| 2 |
| 3 |
1-
| x |
| 3 |
| x |
| 3 |
| 2 |
| 3 |
| x |
| 3 |
1-
| 2 |
| 3 |
| 2 |
| 3 |
| x |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
| x |
| 3 |
| x |
| 3 |
| 1 |
| 3 |
| x |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
x=1
②42:
| 1 |
| 3 |
| 5 |
| 7 |
| 1 |
| 3 |
| 5 |
| 7 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
x=90
点评:本题考查了学生利用等式的性质和比例的基本性质解方程的能力,注意等号对齐.
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