Đk: \(x\le4\)
PT\(\Leftrightarrow\left\{{}\begin{matrix}4-x=\left(x-1\right)^2\\x-1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x-3=0\\x\ge1\end{matrix}\right.\) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{1+\sqrt{13}}{2}\\x=\dfrac{1-\sqrt{13}}{2}\end{matrix}\right.\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{1+\sqrt{13}}{2}\) (tm)
Vậy...
ĐKXĐ: $x \neq 4$
\(\Leftrightarrow\left\{{}\begin{matrix}4-x=x^2-2x+1\\x\ge1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-x-3=0\\x\ge1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(2x-1\right)^2=13\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\sqrt{13}+1}{2}\\x=\dfrac{-\sqrt{13}+1}{2}\end{matrix}\right.\\x>=1\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\sqrt{13}+1}{2}\)
Vậy pt đã cho có tập nghiệm $S={\dfrac{\sqrt[]{13}+1}{2}}$
