Đáp án:
$\begin{array}{l}
Dkxd:R\\
Dat:x\sqrt {2{x^2} + 4} = a\\
\Rightarrow {a^2} = {x^2}.\left( {2{x^2} + 4} \right)\\
\Rightarrow {a^2} = 2.{x^2}\left( {{x^2} + 2} \right)\\
\Rightarrow {x^2}\left( {{x^2} + 2} \right) = \dfrac{{{a^2}}}{2}\\
Pt:{x^2}\left( {{x^2} + 2} \right) = 4 - x\sqrt {2{x^2} + 4} \\
\Rightarrow \dfrac{{{a^2}}}{2} = 4 - a\\
\Rightarrow {a^2} + 2a - 8 = 0\\
\Rightarrow \left( {a - 2} \right)\left( {a + 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a = 2\\
a = - 4
\end{array} \right.\\
+ Khi:a = 2\\
\Rightarrow {x^2}\left( {{x^2} + 2} \right) = \dfrac{{{a^2}}}{2} = 2\\
\Rightarrow {x^4} + 2{x^2} = 2\\
\Rightarrow {\left( {{x^2} + 1} \right)^2} = 3\\
\Rightarrow {x^2} + 1 = \sqrt 3 - 1\\
\Rightarrow {x^2} = \sqrt 3 - 2\left( {vn} \right)\\
+ Khi:a = - 4\\
\Rightarrow {x^2}\left( {{x^2} + 2} \right) = \dfrac{{{a^2}}}{2} = 8\\
\Rightarrow {x^4} + 2{x^2} - 8 = 0\\
\Rightarrow \left( {{x^2} - 2} \right)\left( {{x^2} + 4} \right) = 0\\
\Rightarrow {x^2} = 2\\
\Rightarrow \left[ \begin{array}{l}
x = \sqrt 2 \\
x = - \sqrt 2
\end{array} \right.\\
Vay\,x = \pm \sqrt 2 \,
\end{array}$
