\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-1=1\\2x-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=1\\x=0\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{1}{2},0,1\right\}\)
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