\(\left(x-2\right)^2\left(x-9\right)=0\)
\(=>\left[{}\begin{matrix}\left(x-2\right)^2=0\\x-9=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x-2=0\\x=9\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\)
\(=>x\in\left\{2;9\right\}\)
\(\left(x-2\right)^2\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\)
Vậy \(S=\left\{2;9\right\}\).
\(\left(x-2\right)^2\left(x-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)\left(x-9\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-9=0\)
\(\Leftrightarrow x=2\) hoặc \(x=9\)
Vậy phương trình có tập nghiệm là \(S=\left\{2;9\right\}\)
