\(\left(x^2-9\right)^2-9\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x+3\right)^2-9\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left[\left(x+3\right)^2-9\right]=0\)
\(\Leftrightarrow\left(x-3\right)^2\left[x^2+6x+9-9\right]=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x^2+6x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x\left(x+6\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{0;3;-6\right\}\)
\(\left(x-3\right)^2\left(x+3\right)^2-9\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x+3-3\right)\left(x+3+3\right)=0\Leftrightarrow x=3;x=0;x=-6\)
\(\left(x^2-9\right)^2-9.\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right).\left(x+3\right).\left(x-3\right).\left(x+3\right)-9.\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2.\left[\left(x+3\right)^2-9\right]=0\)
\(\Leftrightarrow\left(x-3\right)^2.\left[\left(x+3-3\right).\left(x+3+3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)^2.x.\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\x=0\\x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\\x=-6\end{matrix}\right.\)
Vậy nghiệm của phương trình là: \(S=\left\{0;3;-6\right\}\)