\(\Leftrightarrow\left(x+2\right)\left(2x-1-x-3\right)=0\)
=>(x+2)(x-4)=0
=>x=-2 hoặc x=4
\(\left(2x-1\right)\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
\(\Leftrightarrow\left(2x-1\right)\left(x+2\right)-\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1-x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
Vậy \(S=\left\{-2;4\right\}\)
⇔(x+2)(2x−1−x−3)=0⇔(x+2)(2x−1−x−3)=0
=>(x+2)(x-4)=0
=>x=-2 hoặc x=4.