<=>
(2x-1 ) ( 3-x ) = 0
=> x = 1/2 hoặc x = 3
3 (2x - 1 ) - x (2x - 1) = 0
=> (2x - 1) (3x - 1) = 0
=> x = 12
hay x = 3
3(2X -1) -X(2X-1) = 0
(2X- 1)(3-X) = 0
2X - 1 = 0 hoặc 3 - X = 0
2X - 1 = 0 \(\Rightarrow\) 2X = 1\(\Rightarrow\) X = 1/2
3 - x = 0 \(\Rightarrow\) X = 3
x \(\in\) {1/2 3)
\(3\left(2x-1\right)-x\left(2x-1\right)=0\)
`<=>(2x-1)(3-x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};3\right\}\)
3 (2x - 1 ) - x (2x - 1) = 0
=> (2x - 1) (3x - 1) = 0
=> x = 12
hay x = 3
