\(\left(x+2\right)^2=64\)
\(\Rightarrow\left[{}\begin{matrix}x+2=8\\x+2=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
\(\left(x+2\right)^2=64\)
\(\Rightarrow\left[{}\begin{matrix}x+2=8\\x+2=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x+2=8\\x+2=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
\(\left(x+2\right)^2=64\)
\(\Rightarrow\left(x+2\right)^2=\left(\pm8\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+2=8\\x+2=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
Vậy x ∈ {6; 10}