\(\left(2x+1\right)^2=49\\ \Rightarrow\left[{}\begin{matrix}2x+1=7\\2x+1=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{3;-4\right\}\)
Ta có: \(\left(2x+1\right)^2=49\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=7\\2x+1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-4\right\}\)
(2x + 1)2 = 49
=> 2x + 1 = 7
=> 2x = 7 - 1 = 6
=> x = 6 : 2 = 3
