Ta có: \(\left(x^2-4x\right)^2+\left(x-2\right)^2=10\)
\(\Leftrightarrow\left(x^2-4x\right)^2+\left(x^2-4x+4\right)=10\)
Đặt \(x^2-4x=t\)
\(\Leftrightarrow t^2+\left(t+4\right)=10\)
\(\Leftrightarrow t^2+t+4-10=0\)
\(\Leftrightarrow t^2+t-6=0\)
\(\Leftrightarrow t^2+3t-2t-6=0\)
\(\Leftrightarrow t\left(t+3\right)-2\left(t+3\right)=0\)
\(\Leftrightarrow\left(t+3\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-4x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\\x^2-4x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x^2-4x+4-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\\left(x-2\right)^2=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x-2=\sqrt{6}\\x-2=-\sqrt{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=2+\sqrt{6}\\x=2-\sqrt{6}\end{matrix}\right.\)
Vậy: \(S=\left\{1;3;2+\sqrt{6};2-\sqrt{6}\right\}\)
