e/ \(x^2-2x+1=4\\↔(x-1)^2-4=0\\↔(x-1-2)(x-1+2)=0\\↔(x-3)(x+1)=0\\↔\left[\begin{array}{1}x-3=0\\x+1=0\end{array}\right.\\↔\left[\begin{array}{1}x=3\\x=-1\end{array}\right.\)
Vậy \(S=\{3;-1\}\)
g/ \(2x^2-3x+1=0\\↔2x^2-2x-x+1=0\\↔2x(x-1)-(x-1)=0\\↔(2x-1)(x-1)=0\\↔\left[\begin{array}{1}2x-1=0\\x-1=0\end{array}\right.\\↔\left[\begin{array}{1}x=\dfrac{1}{2}\\x=1\end{array}\right.\)
Vậy \(S=\{\dfrac{1}{2};1\}\)
h/ \(x^2-5x+6=0\\↔x^2-2x-3x+6=0\\↔x(x-2)-3(x-2)=0\\↔(x-3)(x-2)=0\\↔\left[\begin{array}{1}x-3=0\\x-2=0\end{array}\right.\\↔\left[\begin{array}{1}x=3\\x=2\end{array}\right.\)
Vậy \(S=\{3;2\}\)
