\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)
`@TH1:`
`x+1/2=0`
`x=0-1/2`
`x=-1/2`
`@TH2:`
`2/3-2x=0`
`2x=2/3-0`
`2x=2/3`
`x=2/3:2`
`x=2/3xx1/2`
`x=1/3`
Vậy `x = {1/3 ; -1/2}`
`(x + 1/2).(2/x - 2x) = 0`
`=>` \(\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=0-\dfrac{1}{2}\\2x=\dfrac{2}{3}-0\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\2x=\dfrac{2}{3}\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}:2\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
