Đáp án: $\left[ \begin{array}{l}
x = 1 + \sqrt 5 \\
x = 1 - \sqrt 5
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
2\left( {{x^2} - 2x} \right) + \sqrt {{x^2} - 2x - 3} - 9 = 0\left( 1 \right)\\
\left( {đkxđ:{x^2} - 2x - 3 \ge 0 \Rightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le - 1
\end{array} \right.} \right)\\
\left( 1 \right) \Rightarrow 2\left( {{x^2} - 2x - 3} \right) + \sqrt {{x^2} - 2x - 3} - 3 = 0\\
\Rightarrow 2{\left( {\sqrt {{x^2} - 2x - 3} } \right)^2} + \sqrt {{x^2} - 2x - 3} - 3 = 0\\
\Rightarrow 2{\left( {\sqrt {{x^2} - 2x - 3} } \right)^2} + 3\sqrt {{x^2} - 2x - 3} - 2\sqrt {{x^2} - 2x - 3} - 3 = 0\\
\Rightarrow \left( {2\sqrt {{x^2} - 2x - 3} + 3} \right)\left( {\sqrt {{x^2} - 2x - 3} - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {{x^2} - 2x - 3} = - \frac{3}{2}\left( {ktm} \right)\\
\sqrt {{x^2} - 2x - 3} = 1
\end{array} \right.\\
\Rightarrow {x^2} - 2x - 3 = 1\\
\Rightarrow {x^2} - 2x - 4 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1 + \sqrt 5 \left( {tm} \right)\\
x = 1 - \sqrt 5 \left( {tm} \right)
\end{array} \right.
\end{array}$
