Đáp án:
$\begin{array}{l}
{x^4} + 4{x^3} + 5{x^2} + 4x + 1 = 0\\
+ Khi:x = 0\\
\Rightarrow 1 = 0\left( {ktm} \right)\\
+ Khi:x \ne 0\\
\Rightarrow \dfrac{{{x^4}}}{{{x^2}}} + \dfrac{{4{x^3}}}{{{x^2}}} + \dfrac{{5{x^2}}}{{{x^2}}} + \dfrac{{4x}}{{{x^2}}} + \dfrac{1}{{{x^2}}} = \dfrac{0}{{{x^2}}}\\
\Rightarrow {x^2} + 4x + 5 + \dfrac{4}{x} + \dfrac{1}{{{x^2}}} = 0\\
\Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 4x + \dfrac{4}{x} + 5 = 0\\
\Rightarrow {x^2} + 2.{x^2}.\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^2}}} - 2 + 4.\left( {x + \dfrac{1}{x}} \right) + 5 = 0\\
\Rightarrow {\left( {x + \dfrac{1}{x}} \right)^2} + 4\left( {x + \dfrac{1}{x}} \right) + 3 = 0\\
\text{Đặt}:\left( {x + \dfrac{1}{x}} \right) = a\\
\Rightarrow {a^2} + 4a + 3 = 0\\
\Rightarrow \left( {a + 1} \right)\left( {a + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a + 1 = 0\\
a + 3 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x + \dfrac{1}{x} + 1 = 0\\
x + \dfrac{1}{x} + 3 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + x + 1 = 0\left( {vn} \right)\\
{x^2} + 3x + 1 = 0
\end{array} \right.\\
\Rightarrow {x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{5}{4} = 0\\
\Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{5}{4}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 3 - \sqrt 5 }}{2}\\
x = \dfrac{{ - 3 + \sqrt 5 }}{2}
\end{array} \right.\left( {tmdk} \right)\\
\text{Vậy}\,x = \dfrac{{ - 3 \pm \sqrt 5 }}{2}
\end{array}$
