Question

Answer 1

\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)\\ \Leftrightarrow\left(x+2\right)3x\)

Answer 2

Answer 3

\(\left(2x+1\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1-x-1\right)=0\)

\(\Leftrightarrow2x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{0;-2\right\}\)

Answer 4

S=0,-2