(2x - 1)6 = (2x - 1)8
=> (2x - 1)8 - (2x - 1)6 = 0
=> (2x - 1)6.[(2x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{cases}}\)=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\)=> \(\orbr{\begin{cases}2x=1\\2x-1\in\left\{1;-1\right\}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x\in\left\{2;0\right\}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x\in\left\{1;0\right\}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{2};1;0\right\}\)
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
=>\(\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
=>\(\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
=>\(\left(2x-1\right)^6\left(1-2x+1\right)\left(1+2x-1\right)=0\)
=>\(\left(2x-1\right)^6\left(2-2x\right)2x=0\)
=>(2x-1)6=0 hoặc 2-2x=0 hoặc x=0
+)Với (2x-1)2=6 => x=\(\frac{1}{2}\)
+)Với 2-2x=0 => x=1
Vậy \(x\in\left\{0;\frac{1}{2};1\right\}\)