\(\left(3x-2\right)\left(3x+2\right)-9\left(x+1\right)x=0\\ \Leftrightarrow\left[\left(3x\right)^2-2^2\right]-9x\left(x+1\right)=0\\ \Leftrightarrow\left(6x^2-4\right)\left(9x^2+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x^2-4=0\\9x^2+9=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}6x^2=4\\9x^2=-9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{4}{6}\\x^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{2}{3}\\x^2=-1\left(v\text{ô}.l\text{ý}\right)\end{matrix}\right.\)
\(\left(3x-2\right)\left(3x+2\right)-9\left(x-1\right)x=0\)
\(\Leftrightarrow\left[\left(3x\right)^2-2^2\right]-9x\left(x-1\right)=0\)
\(\Leftrightarrow\left(9x^2-4\right)-9x^2+9x=0\)
\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(9x-4\right)=0\)
\(\Leftrightarrow9x-4=0\)
\(\Leftrightarrow9x=4\)
\(\Leftrightarrow x=\dfrac{4}{9}\)
Vậy: \(x=\dfrac{4}{9}\)