\(x^2-2x-3=0\)
\(\Leftrightarrow x^2-3x-x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy ..
x.(x-2)-3=0
x.(x-2)=0+3
x.(x-2)=3
\(\Rightarrow\left[{}\begin{matrix}x=3\\x-2=3\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
Vậy x=5 hơạc x=3