Ta có: \(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Leftrightarrow\left(3x-8\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\3x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x=8\\3x=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{8}{3};\frac{-2}{3}\right\}\)