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