题目内容
| 解方程 (1)x÷15+0.4=x÷12-0.1 |
(2)150.5-4x+0.5x=7. |
分析:(1)先化简,再根据等式的性质,在方程两边同时减去
x,再加上0.1,最后乘上60求解,
(2)先化简,再根据等式的性质,在方程两边同时加上3.5x,再减去7,最后除以3.5求解.
| 1 |
| 15 |
(2)先化简,再根据等式的性质,在方程两边同时加上3.5x,再减去7,最后除以3.5求解.
解答:解:(1)x÷15+0.4=x÷12-0.1,
x+0.4=
x-0.1,
x+0.4-
x=
x-0.1-
x,
0.4+0.1=
x-0.1+0.1,
0.5×60=
x×60,
x=30;
(2)150.5-4x+0.5x=7,
150.5-3.5x=7,
150.5-3.5x+3.5x=7+3.5x,
150.5-7=7+3.5x-7,
143.5÷3.5=3.5x÷3.5,
x=41.
| 1 |
| 15 |
| 1 |
| 12 |
| 1 |
| 15 |
| 1 |
| 15 |
| 1 |
| 12 |
| 1 |
| 15 |
0.4+0.1=
| 1 |
| 60 |
0.5×60=
| 1 |
| 60 |
x=30;
(2)150.5-4x+0.5x=7,
150.5-3.5x=7,
150.5-3.5x+3.5x=7+3.5x,
150.5-7=7+3.5x-7,
143.5÷3.5=3.5x÷3.5,
x=41.
点评:本题考查了学生根据等式的性质解方程的能力,注意等号对齐.
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