Đáp án:
\[\left[ \begin{array}{l}
x = 0\\
x = - 11 + \sqrt {61}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left( {x + 6} \right)\sqrt {2{x^2} + 1} = {x^2} + x + 6\\
\Rightarrow x > - 6\\
\Leftrightarrow {\left( {x + 6} \right)^2}\left( {2{x^2} + 1} \right) = {\left( {{x^2} + x + 6} \right)^2}\\
\Leftrightarrow \left( {{x^2} + 12x + 36} \right)\left( {2{x^2} + 1} \right) = {x^4} + {x^2} + 36 + 2{x^3} + 12x + 12{x^2}\\
\Leftrightarrow 2{x^4} + {x^2} + 24{x^3} + 12x + 72{x^2} + 36 = {x^4} + 2{x^3} + 13{x^2} + 12x + 36\\
\Leftrightarrow {x^4} + 22{x^3} + 60{x^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
{x^2} + 22x + 60 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 11 + \sqrt {61} \\
x = - 11 - \sqrt {61} \left( L \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 11 + \sqrt {61}
\end{array} \right.
\end{array}\)
