题目内容
| 解方程. 19-120%x=7 |
75x+25=40 | ||||||
8-5%x=
|
x+(0.4-35)=
|
分析:(1)等式的两边同时加上120%x,把原式改写成120%x+7=19,等式的两边同时减去7,然后等式的两边同时除以120%即可;
(2)等式的两边同时减去25,然后等式的两边同时除以75即可;
(3)等式的两边同时加上5%x,把原式改写成5%x+
=8,等式的两边同时减去
,然后等式的两边同时除以5%即可;
(4)等式的两边同时加上35,然后等式的两边同时减去0.4即可.
(2)等式的两边同时减去25,然后等式的两边同时除以75即可;
(3)等式的两边同时加上5%x,把原式改写成5%x+
| 1 |
| 25 |
| 1 |
| 25 |
(4)等式的两边同时加上35,然后等式的两边同时减去0.4即可.
解答:解:
(1)19-120%x=7,
19-120%x+120%x=7+120%x,
120%x+7=19,
120%x+7-7=19-7,
120%x=12,
120%x÷120%=12÷120%,
x=10;
(2)75x+25=40,
75x+25-25=40-25,
75x=15,
75x÷75=15÷75,
x=0.2;
(3)8-5%x=
,
8-5%x+5%x=
+5%x,
5%x+
=8,
5%x+
-
=8-
,
5%x=7.96,
5%x÷5%=7.96÷5%,
x=159.2;
(4)x+(0.4-35)=
×
,
x+0.4-35=0.125,
x+0.4-35+35=0.125+35,
x+0.4=35.125,
x+0.4-0.4=35.125-0.4,
x=34.725.
(1)19-120%x=7,
19-120%x+120%x=7+120%x,
120%x+7=19,
120%x+7-7=19-7,
120%x=12,
120%x÷120%=12÷120%,
x=10;
(2)75x+25=40,
75x+25-25=40-25,
75x=15,
75x÷75=15÷75,
x=0.2;
(3)8-5%x=
| 1 |
| 25 |
8-5%x+5%x=
| 1 |
| 25 |
5%x+
| 1 |
| 25 |
5%x+
| 1 |
| 25 |
| 1 |
| 25 |
| 1 |
| 25 |
5%x=7.96,
5%x÷5%=7.96÷5%,
x=159.2;
(4)x+(0.4-35)=
| 1 |
| 4 |
| 1 |
| 2 |
x+0.4-35=0.125,
x+0.4-35+35=0.125+35,
x+0.4=35.125,
x+0.4-0.4=35.125-0.4,
x=34.725.
点评:本题主要考查解方程,根据等式的性质进行解答即可.
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