\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\)
+) \(\left(2x-15\right)^3=0\Rightarrow2x-15=0\Rightarrow x=7,5\)
+) \(\left(2x-15\right)^2-1=0\Rightarrow\left(2x-15\right)^2=1\)
\(\Rightarrow\left[{}\begin{matrix}2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{7,5;8;7\right\}\)
(2x-15)5 = ( 2x- 15)3
(2x-15)5- (2x-15)3=0
(2x-15)3 . 1-(2x-15)2=0
(2x-15)3.[ 1- (2x-15)2 ] = 0
\(\Rightarrow\) (2x-15)3=0 ( th1)
hoặc 1- (2x-15)2=0 (th2)
(th1) (2x-15)3= 0 (th2) 1-(2x-15)2=0
\(\Rightarrow\)(2x-15)3=03 -(2x-15)2=0-1
2x-15=0 -(2x-15)2=-1
2x =0+15 (2x-15)2=1
2x=15 \(\Rightarrow\) (2x-15)2=12
x=15:2 (2x-15)2=(-1)2
x=7,5 \(\Rightarrow\) 2x-15=1 \(\Rightarrow\) 2x= 1+ 15\(\Rightarrow\) 2x=16
\(\Rightarrow\) 2x-15= -1\(\Rightarrow\) 2x= -1+15\(\Rightarrow\) 2x=14
\(\Rightarrow\) x=16:2 \(\Rightarrow\) x=8
\(\Rightarrow\) x= 14:2\(\Rightarrow\) x=7
Vậy x = 7,5; x=8;x=7