Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{DK:{x^2} - 2x - 1 \ge 0 \Leftrightarrow x \in \left( { - \infty ;1 - \sqrt 2 } \right] \cup \left[ {1 + \sqrt 2 ; + \infty } \right)}\\
{{x^2} - 2x - 1 = \sqrt {({x^2} + 1).(x + 1)} }\\
{ \to {x^4} + 4{x^2} + 1 - 4{x^3} - 2{x^2} + 4x = {x^3} + {x^2} + x + 1}\\
{ \to {x^4} - 5{x^3} + {x^2} + 3x = 0}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{{x^3} - 5{x^2} + x + 3 = 0}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{{x^3} - {x^2} - 4{x^2} + 4x - 3x + 3 = 0}
\end{array}} \right.}\\
{{\rm{\;}} \to \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{{x^2}\left( {x - 1} \right) - 4x\left( {x - 1} \right) - 3\left( {x - 1} \right) = 0}
\end{array}} \right.}\\
{{\rm{\;}} \to \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{\left( {x - 1} \right)\left( {{x^2} - 4x - 3} \right) = 0}
\end{array}} \right.}\\
{{\rm{\;}} \to \left[ {\begin{array}{*{20}{l}}
{x = 0\left( {KTM} \right)}\\
{x = 1\left( {KTM} \right)}\\
{x = 2 \pm 7\left( {TM} \right)}
\end{array}} \right.}
\end{array}\)
