\(\left(2x-1\right)^2-1=35\)
\(\left(2x-1\right)^2=35+1=36\)
\(\Rightarrow2x-1=\pm\sqrt{36}\)
\(2x-1=\pm6\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1=6\Rightarrow2x=6+1=7\Rightarrow x=3,5\\2x-1=-6\Rightarrow2x=-6+1=-5\Rightarrow x=-2,5\end{matrix}\right.\)
Vậy x = \(\left\{3,5;-2,5\right\}\)