Đáp án:
$\begin{array}{l}
2{x^4} + 5{x^3} + {x^2} + 5x + 2 = 0\\
+ Khi:x = 0\\
\Leftrightarrow 2 = 0\left( {ktm} \right)\\
+ Khi:x \ne 0\\
\Leftrightarrow 2{x^2} + 5x + 1 + \dfrac{5}{x} + \dfrac{2}{{{x^2}}} = 0\left( {chia\,{x^2}} \right)\\
\Leftrightarrow 2\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) + 5.\left( {x + \dfrac{1}{x}} \right) + 1 = 0\\
\Leftrightarrow 2.\left( {{x^2} + 2.{x^2}.\dfrac{1}{{{x^2}}} + \dfrac{1}{x}} \right) - 2.2.\dfrac{1}{{{x^2}}}.{x^2} + 5.\left( {x + \dfrac{1}{x}} \right) + 1 = 0\\
\Leftrightarrow 2.{\left( {x + \dfrac{1}{x}} \right)^2} + 5.\left( {x + \dfrac{1}{x}} \right) - 3 = 0\\
Dat:x + \dfrac{1}{x} = a\\
\Leftrightarrow 2{a^2} + 5a - 3 = 0\\
\Leftrightarrow \left( {2a - 1} \right)\left( {a + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a = \dfrac{1}{2}\\
a = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{x} = \dfrac{1}{2}\\
x + \dfrac{1}{x} = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2{x^2} + 2 = x\\
{x^2} + 1 = - 3x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2{x^2} - x + 2 = 0\\
{x^2} + 3x + 1 = 0
\end{array} \right.\\
\Leftrightarrow x = \dfrac{{ - 3 \pm \sqrt 5 }}{2}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{ - 3 \pm \sqrt 5 }}{2}
\end{array}$
