=> 2x-1 \(\in\left\{-1;0;1\right\}\)
=> x \(\in\left\{0;\frac{1}{2};1\right\}\)
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)\right]\left[1+\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left(1-2x+1\right)\left(1+2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left(-2x+2\right).2x=0\)
\(\Leftrightarrow2x-1=0\) hoặc \(-2x+2=0\) hoặc \(2x=0\)
\(\Leftrightarrow2x=1\) hoặc \(-2x=-2\) hoặc \(x=0\)
\(\Leftrightarrow x=\frac{1}{2}\) hoặc \(x=1\) hoặc \(x=0\)
Vậy \(x\in\left\{\frac{1}{2};1;0\right\}\)
