题目内容
(2009?大竹县)求未知数.
(1)
x+
x=
(2)
=
(3)4x+3×0.7=6.5.
(1)
| 1 |
| 3 |
| 2 |
| 5 |
| 4 |
| 15 |
(2)
| 0.25 |
| x |
| 1.25 |
| 3 |
(3)4x+3×0.7=6.5.
分析:(1)先计算出
x+
x=
x,根据等式的性质,两边同乘
即可;
(2)先根据比例的基本性质,把原式转化为1.25x=0.25×3,根据等式的性质,在方程两边同时除以1.25求解;
(3)先求出3×0.7=2.1,根据等式的性质,两边同减去2.1,再同除以4即可.
| 1 |
| 3 |
| 2 |
| 5 |
| 11 |
| 15 |
| 15 |
| 11 |
(2)先根据比例的基本性质,把原式转化为1.25x=0.25×3,根据等式的性质,在方程两边同时除以1.25求解;
(3)先求出3×0.7=2.1,根据等式的性质,两边同减去2.1,再同除以4即可.
解答:解:(1)
x+
x=
x×
=
×
,
x=
;
(2)
=
1.25x=0.25×3,
1.25x÷1.25=0.25×3÷1.25,
x=0.6;
(3)4x+3×0.7=6.5
4x+2.1=6.5,
4x+2.1-2.1=6.5-2.1,
4x÷4=4.4÷4,
x=1.1.
| 1 |
| 3 |
| 2 |
| 5 |
| 4 |
| 15 |
| 11 |
| 15 |
| 15 |
| 11 |
| 4 |
| 15 |
| 15 |
| 11 |
x=
| 4 |
| 11 |
(2)
| 0.25 |
| x |
| 1.25 |
| 3 |
1.25x=0.25×3,
1.25x÷1.25=0.25×3÷1.25,
x=0.6;
(3)4x+3×0.7=6.5
4x+2.1=6.5,
4x+2.1-2.1=6.5-2.1,
4x÷4=4.4÷4,
x=1.1.
点评:本题主要考查了学生根据比例的基本性质和等式的性质解方程的能力,注意等号对齐.
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