\(\left(x-\dfrac{2}{3}\right)\left(x+\dfrac{1}{4}\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\x+\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy ...
\(\left(x-\dfrac{2}{3}\right)\left(x+\dfrac{1}{4}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=0\\x+\dfrac{1}{4}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{-1}{4}\end{matrix}\right.\)
Vậy \(x=\dfrac{2}{3}\) hoặc \(x=\dfrac{-1}{4}\)