\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=\pm1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy.........
Ta có: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-15-1\right)\left(2x-15+1\right)=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{15}{2};8;7\right\}\)