\(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow2-x-2\left(x-2\right)^3=0\)
\(\Leftrightarrow-\left(x-2\right)-2\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)\left[-1-2\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[-1-2\left(x^2-4x+4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(-1-2x^2+4x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x^2+4x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-2x^2+4x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\-2\left(x^2-2x+1\right)-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\-2\left(x^2-2x+1\right)-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\-2\left(x-1\right)^2=3\left(loai\right)\end{matrix}\right.\)
Vậy x=2
is ............