\(x^2-9=3\left(x-3\right)\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}\)
\(x^2-9=3\left(x-3\right)\\ \Leftrightarrow\left(x-3\right)^2-3\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)-3\left(x-3\right)\\ \Leftrightarrow\left(x-3\right)\left(x+3-3\right)=0\\ \Leftrightarrow\left(x-3\right)x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy phương trình có 2 nghiệm là \(x=0,x=3\)
\(x^2-9=3\left(x-3\right)\)
\(x^2-9-3\left(x-3\right)=0\)
\(x^2-9-3x+9=0\)
\(x^2-3x=0\)
\(x\left(x-3\right)=0\)
\(x=0\) hoặc \(x-3=0\)
*) \(x-3=0\)
\(x=0+3\)
\(x=3\)
Vậy \(x=0\); \(x=3\)
