Đáp án:
$\begin{array}{l}
\left( {x + 2} \right)\sqrt {{x^2} + 2} = {x^2} + 3x + 1\\
\Rightarrow {\left( {x + 2} \right)^2}\left( {{x^2} + 2} \right) = {\left( {{x^2} + 3x + 1} \right)^2}\\
\Rightarrow \left( {{x^2} + 4x + 4} \right)\left( {{x^2} + 2} \right)\\
= {x^4} + 9{x^2} + 1 + 2.{x^2}.3x + 2.{x^2}.1 + 2.3x.1\\
\Rightarrow {x^4} + 2{x^2} + 4{x^3} + 8x + 4{x^2} + 8 = {x^4} + 6{x^3} + 11{x^2} + 6x + 1\\
\Rightarrow 2{x^3} + 5{x^2} - 2x - 7 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1,13\\
x = - 1,37\\
x = - 2,25
\end{array} \right.
\end{array}$
