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\(\left(5x-1\right)\left(x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}VP:5x-1=0\\VT:x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{3}\end{cases}}\)
Vậy x của VP là 1/5 và x của VT là 1/3
(5x - 1).(x - 1/3) = 0
<=> \(\orbr{\begin{cases}5x-1=0\\x-\frac{1}{3}=0\end{cases}}\)
+) 5x -1 = 0 => x =1/5
+) x - 1/3 = 0 => x=1/3
Vậy x=1/5 , x=1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
(5x - 1)(2x - 1/3) = 0
<=> 5x - 1 = 0 hoặc 2x - 1/3 = 0
=> x = 1/5 hoặc x = 1/6
vậy x= 1/5 hoặc x= 1/6
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a) (5x - 1) . ( 2x - 1/3 ) = 0
=> 5x - 1 = 0
2x - 1/3 = 0
=> 5x = 1
2x = 1/3
=> x = 1/5
x = 1/6
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(5x-1\right)\left(\frac{2x-1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{2}\end{cases}}\)
b) \(6\left(x-1\right)+2x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\6+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
a) \(\left(5x-1\right)\cdot\frac{2x-1}{3}=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\\frac{2x-1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=1\\2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(x=\frac{1}{5};x=\frac{1}{2}\)
b) 6(x-1)+2x(x-1)=0
<=> (x-1)(6+2x)=0
<=> \(\orbr{\begin{cases}x-1=0\\6+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy x=1; x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\hept{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
|5\(x\) - 4| = |\(x+2\)|
\(\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
vậy \(x\in\) { \(\dfrac{1}{3};\dfrac{3}{2}\)}
|2\(x\) - 3| - |3\(x\) + 2| = 0
|2\(x\) - 3| = | 3\(x\) + 2|
\(\left[{}\begin{matrix}2x-3=3x+2\\2x-3=-3x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{5}\end{matrix}\right.\)
vậy \(x\in\){ -5; \(\dfrac{1}{5}\)}
![](https://rs.olm.vn/images/avt/0.png?1311)
(5x-1).(2x-1/3)=0
(5x.2x)-(1+1/3)=0
10x-4/3=0
Vậy 10x=4/3
10x=4/3
x=4/3:10
x=2/15
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
\(5x\left(x-\frac{1}{3}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)
`3-1/5x=0`
`=> 1/5x=3-0`
`=> 1/5x=3`
`=> x= 3 : 1/5`
`=> x= 3 xx 5`
`=> x= 15`
3 - \(\dfrac{1}{5}\) \(x\) = 0
\(\dfrac{1}{5}x\) = 3
\(x\) = 3 : \(\dfrac{1}{5}\)
\(x\) = 15