Có: \(x^2-2x-1=\left(x^2-2x+1\right)-2\)
\(=\left(x-1\right)^2-2\)
Đặt \(\left(x-1\right)^2-2=0\)
\(\rightarrow\left(x-1\right)^2=2\)
\(\rightarrow\left\{{}\begin{matrix}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1+\sqrt{2};1-\sqrt{2}\right\}\)
Ta có : \(x^2-2x-1=x^2-2x+1-2\)\(=\left(x-1\right)^2-\left(\sqrt{2}\right)^2\)
\(=\left(x-1-\sqrt{2}\right)\left(x-1+\sqrt{2}\right)\)
Cho \(x^2-2x-1=0\)
\(\Leftrightarrow\left(x-1-\sqrt{2}\right)\left(x-1+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1-\sqrt{2}=0\\x-1+\sqrt{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\)
Vậy nghiệm của đa thức là \(x=1-\sqrt{2};x=1+\sqrt{2}\)
