Đáp án:
\[\left[ \begin{array}{l}
x = 6\\
x = - 2
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^2} - 4x - 6 = \sqrt {2{x^2} - 8x + 12} \\
\Leftrightarrow {x^2} - 4x - 6 - \sqrt {2{x^2} - 8x + 12} = 0\\
\Leftrightarrow 2{x^2} - 8x - 12 - 2\sqrt {2{x^2} - 8x + 12} = 0\\
\Leftrightarrow \left( {2{x^2} - 8x + 12} \right) - 2\sqrt {2{x^2} - 8x + 12} - 24 = 0\\
\Leftrightarrow \left[ {\left( {2{x^2} - 8x + 12} \right) - 6\sqrt {2{x^2} - 8x + 12} } \right] + \left[ {4\sqrt {2{x^2} - 8x + 12} - 24} \right] = 0\\
\Leftrightarrow \sqrt {2{x^2} - 8x + 12} \left( {\sqrt {2{x^2} - 8x + 12} - 6} \right) + 4.\left( {\sqrt {2{x^2} - 8x + 12} - 6} \right) = 0\\
\Leftrightarrow \left( {\sqrt {2{x^2} - 8x + 12} - 6} \right)\left( {\sqrt {2{x^2} - 8x + 12} + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {2{x^2} - 8x + 12} - 6 = 0\\
\sqrt {2{x^2} - 8x + 12} + 4 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {2{x^2} - 8x + 12} = 6\\
\sqrt {2{x^2} - 8x + 12} = - 4
\end{array} \right.\\
\Leftrightarrow \sqrt {2{x^2} - 8x + 12} = 6\\
\Leftrightarrow 2{x^2} - 8x + 12 = 36\\
\Leftrightarrow 2{x^2} - 8x - 24 = 0\\
\Leftrightarrow {x^2} - 4x - 12 = 0\\
\Leftrightarrow \left( {x - 6} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = - 2
\end{array} \right.
\end{array}\)
