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