ta có \(\left(x+2\right)^3-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)^3=9\left(x+2\right)\\ \Leftrightarrow\left(x+2\right)^2=9\\ \Rightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
vậy \(x\in\left\{-5;1\right\}\)
\(\left(x+2\right)^3-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2=9\)(Chuyển vế rồi chia cả hai vế cho(x+2))
\(\Leftrightarrow x+2=3\) hoặc \(x+2=-3\)
\(\Leftrightarrow x=1\) hoặc \(x=-5\)
Vậy .......