a) \(\left(2x+1\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
b) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-2+1\)
\(\Rightarrow x=\dfrac{-2+1}{2}\)
\(\Rightarrow x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
\(a.\left(2x+1\right)^2=25\)
\(\Leftrightarrow\left(2x+1\right)^2-25=0\)
\(\Leftrightarrow\left(2x+1+5\right)\left(2x+1-5\right)=0\)
\(\Leftrightarrow\left(2x+6\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+6=0\\2x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-3;2\right\}\)
