Đáp án:
$x \in \left\{ { \pm \sqrt 3 ;\,\,\, \pm 2\sqrt 6 } \right\}.$
Giải thích các bước giải:
\(\begin{array}{l}
{x^2} + 11 = 7\sqrt {{x^2} + 1} \Leftrightarrow {x^2} + 1 - 7\sqrt {{x^2} + 1} + 10 = 0\,\,\,\,\,\left( * \right)\\
Dat\,\,\,\sqrt {{x^2} + 1} = t\,\,\,\left( {t \ge 0} \right)\\
\Rightarrow \left( * \right) \Leftrightarrow {t^2} - 7t + 10 = 0\\
\Leftrightarrow {t^2} - 2t - 5t + 10 = 0\\
\Leftrightarrow t\left( {t - 2} \right) - 5\left( {t - 2} \right) = 0\\
\Leftrightarrow \left( {t - 2} \right)\left( {t - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 2\,\,\left( {tm} \right)\\
t = 5\,\,\,\left( {tm} \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt {{x^2} + 1} = 2\\
\sqrt {{x^2} + 1} = 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 1 = 4\\
{x^2} + 1 = 25
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 3\\
{x^2} = 24
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \sqrt 3 \\
x = \pm 2\sqrt 6
\end{array} \right..\\
\Rightarrow x \in \left\{ { \pm \sqrt 3 ;\,\,\, \pm 2\sqrt 6 } \right\}.
\end{array}\)
