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![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x:
b) 1/3.x+2/5.(x-1)=0
\(<=> \dfrac{1}{3} .x +\dfrac{2}{5}x - \dfrac{2}{5} =0\)
\(<=> \dfrac{11}{15}x = \dfrac{2}{5}\)
\(<=> x= \dfrac{6}{11}\)
Vậy \( x= \dfrac{6}{11}\)
c) (2x-3).(6-2x)=0
\(<=> \begin{cases}
2x-3=0 \\
6-2x=0
\end{cases}\) \(<=> \begin{cases}
2x=3 \\
-2x=-6
\end{cases}\) \(<=>\begin{cases}
x=\dfrac{3}{2} \\
x=3
\end{cases}\)
Vậy \(x=( \dfrac{3}{2} ; 3)\)
d) -2/3-1/3.(2x-5)= 3/2
\(<=> 2x-5= \dfrac{5}{2}\)
\(<=> 2x= \dfrac{15}{2}\)
\(<=> x= \dfrac{15}{4}\)
Vậy \(x= \dfrac{15}{4}\)
f) 1/3.x-1/2=4 và 1/2 (Hỗn số ý '^')
\(<=> \dfrac{1}{3} x -\dfrac{1}{2} = \dfrac{9}{2}\)
\(<=> \dfrac{1}{3}x =5\)
\(<=> x= 15\)
Vậy \(x= 15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a; - \(\dfrac{1}{3}\).(15\(x-9\)) + \(\dfrac{2}{7}\).(- \(x-34\)) = 1 - \(\dfrac{3}{4}\).(-16\(x+4\))
- 5\(x\) + 3 - \(\dfrac{2}{7}\)\(x\) - \(\dfrac{68}{7}\) = 1 + 12\(x\) - 3
12\(x\) + 5\(x\) + \(\dfrac{2}{7}x\) = 3 - \(\dfrac{68}{7}\) - 1 + 3
17\(x\) + \(\dfrac{2}{7}x\) = (3 - 1 + 3) - \(\dfrac{68}{7}\)
\(\dfrac{121}{7}\)\(x\) = 5 - \(\dfrac{68}{7}\)
\(\dfrac{121}{7}\) \(x\) = - \(\dfrac{33}{7}\)
\(x\) = - \(\dfrac{33}{7}\): \(\dfrac{121}{7}\)
\(x\) = - \(\dfrac{3}{11}\)
Vậy \(x\) = - \(\dfrac{3}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
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Giúp tớ với tớ cần ngay bây giờ![gianroi gianroi](https://hoc24.vn/media/cke24/plugins/smiley/images/gianroi.png)
![gianroi gianroi](https://hoc24.vn/media/cke24/plugins/smiley/images/gianroi.png)
a. \(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)-\dfrac{3}{2}=\dfrac{1}{4}\\ \Leftrightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \Leftrightarrow\dfrac{1}{2}x=\dfrac{29}{24}\\ \Leftrightarrow x=\dfrac{29}{12}\)
b. \(\dfrac{3}{4}-2\left(2x-\dfrac{2}{3}\right)=2\\ \Leftrightarrow2\left(2x-\dfrac{2}{3}\right)=-\dfrac{5}{4}\\ \Leftrightarrow2x-\dfrac{2}{3}=-\dfrac{5}{8}\\ \Leftrightarrow2x=\dfrac{1}{24}\\ \Leftrightarrow x=\dfrac{1}{48}\)
c. \(\left(2x-3\right)\left(6-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
Vậy ..........