a) \(\left(2x-3\right)^2=25\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm5\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)
Vậy: x = 4 hoặc x = -1
b) \(\frac{27}{3x}=3\Leftrightarrow\frac{9}{x}=3\Leftrightarrow9=3x\Leftrightarrow3x=9\Leftrightarrow x=\frac{9}{3}=3\)
=> x = 3