Đặt: \(x^2-2x+1=t\left(t\ge0\right)\)
pt <=> \(\left(t+1\right)\left(t-2\right)=2\)
\(\Leftrightarrow t^2-t-4=0\)
\(\Leftrightarrow\left(t^2-2\cdot\dfrac{1}{2}t+\dfrac{1}{4}\right)-\dfrac{17}{4}=0\)
\(\Leftrightarrow\left(t-\dfrac{1}{2}\right)^2=\dfrac{17}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}t-\dfrac{1}{2}=\dfrac{\sqrt{17}}{2}\\t-\dfrac{1}{2}=-\dfrac{\sqrt{17}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{1+\sqrt{17}}{2}\left(tm\right)\\t=\dfrac{1-\sqrt{17}}{2}\left(ktm\right)\end{matrix}\right.\)
Ta có: \(t=\dfrac{1+\sqrt{17}}{2}\)
\(\Rightarrow x^2-2x+1=\dfrac{1+\sqrt{17}}{2}\)
\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1+\sqrt{17}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{\dfrac{1+\sqrt{17}}{2}}\\x-1=-\sqrt{\dfrac{1+\sqrt{17}}{2}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{\dfrac{1+\sqrt{17}}{2}}\\x=1-\sqrt{\dfrac{1+\sqrt{17}}{2}}\end{matrix}\right.\)
Vậy pt có 2 nghiệm........................
