\(\left(x+4\right)\left(9x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\9x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x^2=\dfrac{1}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\pm\dfrac{1}{3}\end{matrix}\right.\)\(S=\left\{-4,\pm\dfrac{1}{3}\right\}\)
Ta có: \(\left(x+4\right)\left(9x^2-1\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\3x=1\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-4;\dfrac{1}{3};-\dfrac{1}{3}\right\}\)