\(\left(2x-3\right)^2=16\)
\(\Rightarrow\left(2x-3\right)^2=4^2\)
\(\Rightarrow2x-3=4\)
\(\Rightarrow2x=4+3\)
\(\Rightarrow2x=7\)
\(\Rightarrow x=\frac{7}{2}\)
\(\left(2x-3\right)^2=16\)
\(\Leftrightarrow\left(2x-3\right)^2=4^2\)
\(\Leftrightarrow2x-3=4\)
\(\Leftrightarrow2x=4+3=7\)
\(\Leftrightarrow x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}\)