Ta có : \(\left(2.x-1\right)^2=3^2.5^2\)
\(\Leftrightarrow\left(2.x-1\right)^2=\left(3.5\right)^2\)
\(\Leftrightarrow\left(2.x-1\right)^2=15^2\)
\(\Leftrightarrow2.x-1=15\)
\(\Leftrightarrow2.x=15+1\)
\(\Leftrightarrow2.x=16\)
\(\Leftrightarrow x=16:2\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
\(\left(2x-1\right)^2=3^2.5^2\)
\(\left(2x-1\right)^2=225\)
\(\left(2x-1\right)^2=\left(\pm15\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=15\\2x-1=-15\end{cases}\Rightarrow\orbr{\begin{cases}2x=16\\2x=-14\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=-7\end{cases}}}\)
\(\Rightarrow x\in\left\{-7;8\right\}\)
