\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3.\left(2x-16\right).\left(2x-14\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=15\\2x=16\\x=14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
`(2x-15)^5 =(2x-15)^3`
`<=> (2x-15)^5 -(2x-15)^3 =0`
`<=>(2x-15)^3*[(2x-15)^2-1]=0`
`@TH1 : (2x-15)^3=0 => 2x-15 =0`
`=> 2x=15 => x=15/2`
`@TH2 : (2x-15)^2 -1 =0`
`=> [(2x-15=1),(2x-15=-1):}`
`=> [(2x=1+15=16),(2x=-1+15=14):}`
`=>[(x=8),(x=7):}`
Vậy `x in {15/2;8;7}`