\(\left(3x+\dfrac{3}{5}\right)\left(\left|x\right|-\dfrac{1}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{3}{5}=0\\\left|x\right|=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{1}{5};\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
⇒\(\left\{{}\begin{matrix}3x+\dfrac{3}{5}=0\\\left|x\right|-\dfrac{1}{4}=0\end{matrix}\right.\) ⇒\(\left\{{}\begin{matrix}3x=0-\dfrac{3}{5}=-\dfrac{3}{5}\\\left|x\right|=0+\dfrac{1}{4}=\dfrac{1}{4}\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x=-\dfrac{3}{5}:3=-\dfrac{1}{5}\\x=\dfrac{1}{4},-\dfrac{1}{4}\end{matrix}\right.\)
\(\left(3x+\dfrac{3}{5}\right).\left(\left|x\right|-\dfrac{1}{4}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+\dfrac{3}{5}=0\\\left|x\right|-\dfrac{1}{4}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=\dfrac{1}{4};x=\dfrac{-1}{4}\end{matrix}\right.\)