=>(x-3)(x-1)=0
=>x=3 hoặc x=1
\(x\left(x-3\right)-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{3;1\right\}\)
x ( x - 3 ) - ( x - 3 ) = 0
( x - 3 ) ( x - 1 ) = 0
\(\left\{{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(x\left(x-3\right)-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)