题目内容
求未知数χ.
χ-
χ=3
(χ+2.1)÷3=6
χ:
=14:
.
解:(1)χ-
χ=3,
x=3,
x
=3
,
x=8;
(2)(χ+2.1)÷3=6,
(χ+2.1)÷3×3=6×3,
x+2.1-2.1=18-2.1,
x=15.9;
(3)χ:
=14:
,
x=
×14,
x
=4
,
x=10.
分析:(1)先化简方程,再依据等式的性质,方程两边同时除以
求解,
(2)依据等式的性质,方程两边同时乘3,再同时减2.1求解,
(3)先依据比例基本性质,化简方程,再依据等式的性质方程两边同时除以
求解.
点评:本题主要考查学生运用等式的性质,以及比例基本性质解方程的能力.
![](http://thumb.1010pic.com/pic5/latex/5192.png)
![](http://thumb.1010pic.com/pic5/latex/597.png)
![](http://thumb.1010pic.com/pic5/latex/597.png)
![](http://thumb.1010pic.com/pic5/latex/214949.png)
![](http://thumb.1010pic.com/pic5/latex/214949.png)
x=8;
(2)(χ+2.1)÷3=6,
(χ+2.1)÷3×3=6×3,
x+2.1-2.1=18-2.1,
x=15.9;
(3)χ:
![](http://thumb.1010pic.com/pic5/latex/35.png)
![](http://thumb.1010pic.com/pic5/latex/335.png)
![](http://thumb.1010pic.com/pic5/latex/335.png)
![](http://thumb.1010pic.com/pic5/latex/35.png)
![](http://thumb.1010pic.com/pic5/latex/335.png)
![](http://thumb.1010pic.com/pic5/latex/157724.png)
![](http://thumb.1010pic.com/pic5/latex/157724.png)
x=10.
分析:(1)先化简方程,再依据等式的性质,方程两边同时除以
![](http://thumb.1010pic.com/pic5/latex/597.png)
(2)依据等式的性质,方程两边同时乘3,再同时减2.1求解,
(3)先依据比例基本性质,化简方程,再依据等式的性质方程两边同时除以
![](http://thumb.1010pic.com/pic5/latex/335.png)
点评:本题主要考查学生运用等式的性质,以及比例基本性质解方程的能力.
![](http://thumb2018.1010pic.com/images/loading.gif)
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