\(\sqrt{\left(x-3\right)^2}=9\\ \Leftrightarrow x-3=9\\ \Leftrightarrow x=12\)
\(\sqrt{\left(x-3\right)^2}=9\)
\(\Rightarrow\sqrt{\left(x-3\right)^2}=\left(\sqrt{9}\right)^2\)
\(\Rightarrow\sqrt{\left(x-3\right)^2}=\sqrt{81}\)
\(\Rightarrow\left(x-3\right)^2=81\)
\(\Rightarrow\left(x-3\right)^2=\left(\pm9\right)^2\)
\(\Rightarrow x-3=\pm9.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9+3\\x=\left(-9\right)+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
Vậy \(x\in\left\{12;-6\right\}.\)
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