\(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\\ \Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x+1\right)\left[\left(3x-2\right)-\left(x-1\right)\right]=0\\ \Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-\frac{2}{3};-1;\frac{1}{2}\right\}\)
