(2x-1)^6=(2x-1)^8'
=> (2x-1)=1 hoặc (2x-1)=0
(2x-1)=1 =>2x=2 =>x=1
(2x-1)=0 =>2x=1 =>x=0,5
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\left(2x-1\right)^6-\left(2x-1\right)^6.\left(2x-1\right)^2=0\)
\(\left(2x-1\right)^6.\left[1-\left(2x-1\right)^2\right]=0\)
\(\left(2x-1\right)^6.\left[\left(1-2x+1\right)\left(1+2x-1\right)\right]=0\)
\(\left(2x-1\right)^6.\left[\left(2-2x\right).2x\right]=0\)
\(\left(2x-1\right)^6.\left[4x\left(1-x\right)\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\4x\left(1-x\right)=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x-1=0\\4x=0hoac1-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0hoacx=1\end{cases}}\)
vay \(x=\frac{1}{2}\)hoac \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(\frac{\left(2-3x\right)^4}{\left(3x-2\right)^5}=\frac{-\left(3x-2\right)^4}{\left(3x-2\right)^5}=\frac{-1}{3x-2}\)