`~rai~`
\(a)\sqrt{x^2-2x+4}=2x-2\\\Leftrightarrow \begin{cases}2x-2\ge 0\\\left(\sqrt{x^2-2x+4}\right)^2=(2x-2)^2\end{cases}\\\Leftrightarrow \begin{cases}x\ge 1\\x^2-2x+4=4x^2-8x+4\end{cases}\\\Leftrightarrow \begin{cases}x\ge 1\\3x^2-6x=0\end{cases}\\\Leftrightarrow \begin{cases}x\ge 1\\3x(x-2)=0\end{cases}\\\Leftrightarrow \begin{cases}x\ge 1\\\left[\begin{array}{I}3x=0\\x-2=0\end{array}\right.\end{cases}\\\Leftrightarrow \begin{cases}x\ge 1\\\left[\begin{array}{I}x=0\\x=2\end{array}\right.\end{cases}\\\Leftrightarrow x=2.\\\text{Vậy phương trình có tập nghiệm là S=}\{2\}.\\b)\sqrt{4x^2-4x+1}=x-1\\\Leftrightarrow\\\begin{cases}x-1\ge 0\\\left(\sqrt{4x^2-4x+1}\right)^2=(x-1)^2\end{cases}\\\Leftrightarrow \begin{cases}x\ge 1\\4x^2-4x+1=x^2-2x+1\end{cases}\\\Leftrightarrow \begin{cases}x\ge 1\\3x^2-2x=0\end{cases}\\\Leftrightarrow \begin{cases}x\ge 1\\x(3x-2)=0\end{cases}\\\Leftrightarrow \begin{cases}x\ge 1\\\left[\begin{array}{I}x=0\\3x-2=0\end{array}\right.\end{cases}\\\Leftrightarrow\begin{cases}x\ge 1\\\left[\begin{array}{I}x=0\\x=\dfrac{2}{3}\end{array}\right.\end{cases}\\\Leftrightarrow x=\varnothing.\\\text{Vậy S=}\varnothing.\)
