题目内容
解比例.
|
X:
| ||||||||
|
0.8:3
|
分析:(1)根据比例的基本性质,把原式转化为1.2X=7.5×0.4,再根据等式的性质,在方程两边同时除以1.2求解,
(2)根据比例的基本性质,把原式转化为10X=60×
,再根据等式的性质,在方程两边同时除以10求解,
(3)根据比例的基本性质,把原式转化为
×(4-X)=
×
,再根据等式的性质,在方程两边同时乘2,再加X,最后减去
求解,
(4)根据比例的基本性质,把原式转化为3
×(5-X)=0.8,再根据等式的性质,在方程两边同时除以3
,再加X,再
求解,
(2)根据比例的基本性质,把原式转化为10X=60×
| 5 |
| 12 |
(3)根据比例的基本性质,把原式转化为
| 1 |
| 2 |
| 2 |
| 7 |
| 1 |
| 3 |
| 4 |
| 21 |
(4)根据比例的基本性质,把原式转化为3
| 3 |
| 4 |
| 3 |
| 4 |
| 16 |
| 75 |
解答:解:(1)
=
,
1.2X=7.5×0.4,
1.2X÷1.2=3÷1.2,
X=2.5;
(2)X:
=60:10,
10X=60×
,
10X÷=25÷10,
X=2.5;
(3)
:
=
:(4-X),
×(4-X)=
×
,
×(4-X)×2=
×2,
4-X+X=
+X,
4-
=
+X-
,
X=3
;
(4)0.8:3
=(5-X),
3
×(5-X)=0.8,
3
×(5-X)÷3
=0.8÷3
,
5-X+X=
+X,
5-
=
+X-
,
X=4
.
| 1.2 |
| 7.5 |
| 0.4 |
| X |
1.2X=7.5×0.4,
1.2X÷1.2=3÷1.2,
X=2.5;
(2)X:
| 5 |
| 12 |
10X=60×
| 5 |
| 12 |
10X÷=25÷10,
X=2.5;
(3)
| 1 |
| 2 |
| 2 |
| 7 |
| 1 |
| 3 |
| 1 |
| 2 |
| 2 |
| 7 |
| 1 |
| 3 |
| 1 |
| 2 |
| 2 |
| 21 |
4-X+X=
| 4 |
| 21 |
4-
| 4 |
| 21 |
| 4 |
| 21 |
| 4 |
| 21 |
X=3
| 17 |
| 21 |
(4)0.8:3
| 3 |
| 4 |
3
| 3 |
| 4 |
3
| 3 |
| 4 |
| 3 |
| 4 |
| 3 |
| 4 |
5-X+X=
| 16 |
| 75 |
5-
| 16 |
| 75 |
| 16 |
| 75 |
| 16 |
| 75 |
X=4
| 59 |
| 75 |
点评:本题主要考查了学生根据等式的性质和比例的基本发生解方程的能力,注意等号对齐.
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