a) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(2x-1\right)^6=0\\2x-1=1\\2x-1=-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=1\\x=0\end{cases}}\)
phần b chuyển vế, đạt nhân tử chung....... làm tương tự phần a
a, Ta có :
\(\left(2x-1\right)^6=\left(2x-1\right)^8\) \(=\left(2x-1\right)^8-\left(2x-1\right)^6\) \(=\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]\) = 0
\(\Rightarrow\orbr{\begin{cases}\left(2x-1\right)^2-1=0\\\left(2x-1\right)^6=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}\left(2x-1\right)^2=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}}\\2x=1\end{cases}}}\)=> \(2x-1=0\) hoặc \(2x-1=-1\) hoặc \(2x-1=1\)
=> \(x=\frac{1}{2};x=0\) hoặc \(x=1\)
Vậy \(x=\frac{1}{2};x=0\) hoặc x = 1
