\(\left(x+\dfrac{1}{2}\right)\left(x-\dfrac{3}{5}\right)=0\\ TH1:x+\dfrac{1}{2}=0\\ =>x=-\dfrac{1}{2}\\ TH2:x-\dfrac{3}{5}=-0\\ =>x=\dfrac{3}{5}\)
\(\left(x+\dfrac{1}{2}\right).\left(x-\dfrac{3}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x-\dfrac{3}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-\dfrac{1}{2}\\x=0+\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{1}{2};\dfrac{3}{5}\right\}\)