\(\left(x^2-4x\right)^2+2\left(x-2\right)^2=43\)
\(\Leftrightarrow\left(x^2-4x\right)^2+2\left(x^2-4x+4\right)=43\)
Đặt \(t=x^2-4x\) ta được:
\(t^2+2\left(t+4\right)=43\)
\(\Leftrightarrow t^2+2t+8=43\Leftrightarrow t^2+2t-35=0\)
\(\Leftrightarrow\left(t-5\right)\left(t+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-5=0\\\\t+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=5\\\\t=-7\end{matrix}\right.\)
Xét t = 5:
\(x^2-4x=5\Leftrightarrow x^2-4x-5=0\Leftrightarrow\left(x+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\\\x=5\end{matrix}\right.\)
Xét t = -7:
\(x^2-4x=-7\Leftrightarrow x^2-4x+7=0\Leftrightarrow\left(x-2\right)^2+3=0\left(vl\right)\)
Vậy, \(S=\left\{-1;5\right\}\)
