\(\left(4x-1\right)^2=\left(1-4x\right)^4\)
\(\Rightarrow\left(4x-1\right)^2=\left(4x-1\right)^4\)
\(\Rightarrow\left(4x-1\right)^2\left[1-\left(4x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(4x-1\right)^2=0\\1-\left(4x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x-1=0\\\left(4x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=1\\4x-1\in\left\{1;-1\right\}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x\in\left\{\dfrac{1}{2};0\right\}\end{matrix}\right.\)
Ta có: \(\left(4x-1\right)^2=\left(1-4x\right)^4\)
Đặt \(\left(4x-1\right)^2=t\) \(\left(t\ge0\right)\)
\(\Rightarrow1-4x=-t\)
\(\Leftrightarrow t=-t^2\)
\(\Leftrightarrow t^2+t=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=-1\left(\text{loại}\right)\end{matrix}\right.\)
Ta có: \(t=0\)
\(\Leftrightarrow(4x-1)^2 =0\)
\(\Leftrightarrow4x-1=0\)
\(\Rightarrow x=\dfrac{1}{4}\)
