\(x^2-2x+3=2\sqrt{2x^2-4x+3}\left(x\in R\right)\)
\(\Leftrightarrow x^2-2x+3=2\sqrt{2x^2-4x+6-3}\)
\(\Leftrightarrow x^2-2x+3=2\sqrt{2\left(x^2-2x+3\right)-3}\)
Đặt: \(t=x^2-2x+3\)
Phương trình trở thành:
\(\Rightarrow t=2\sqrt{2t-3}\) \(\left(t\ge\dfrac{3}{2}\right)\)
\(\Leftrightarrow t^2=4\left(2t-3\right)\)
\(\Leftrightarrow t^2=8t-12\)
\(\Leftrightarrow t^2-8t+12=0\)
\(\Leftrightarrow\left(t-2\right)\left(t-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=6\end{matrix}\right.\) (tm)
+) Với \(t=2\)
\(\Leftrightarrow x^2-2x+3=2\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
+) Với \(t=6\)
\(\Leftrightarrow x^2-2x+3=6\)
\(\Leftrightarrow x^2-2x+3-6=0\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{1;-1;3\right\}\)
