\(x\left(x+3\right)=\left(2x+1\right)\left(x+3\right)\)
\(\Leftrightarrow x\left(x+3\right)-\left(2x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
Vậy.........
\(x\left(x+3\right)=\left(2x+1\right)\left(x+3\right)\)
\(\Leftrightarrow x\left(x+3\right)-\left(2x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[x-\left(2x+1\right)\right]\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-2x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2x+1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+1=0\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Tập nghiệm của pt là : \(S=\left\{1;-1\right\}\)
x(x+3)=(2x+1)(x+3)
<=> x(x+3)-(2x+1)(x+3)=0
<=> (x+3)(x-2x-1)=0
<=>(x+3)(-x-1)=0
<=>[x+3=0 <=>[x=-3
[-x-1=0. [x =-1
Vậy pt đã cho có tập nghiệm S={-3;-1}