(x-3)^2 - (x-3)(2x+1) =0
⇔ ( x-3) ( x - 3 - 2x -1 ) =0
⇔ ( x - 3 ) ( -4 -x ) =0
⇔\(\left[{}\begin{matrix}x-3=0\\-4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
\(\left(x-3\right)^2-\left(x-3\right).\left(2x+1\right)=0\)
\(\Rightarrow\left(x-3\right).\left(x-3-2x-1\right)=0\)
\(\Rightarrow\left(x-3\right).\left(-1x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\-1x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+3\\-1x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=4:\left(-1\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{3;-4\right\}.\)
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