Đáp án:

 Hệ phương trình đã cho vô nghiệm.

Giải thích các bước giải:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^4} + 4 - 5x\left( {{x^2} - 2x + 2} \right) < 0\\
{x^4} - 1 > 4x\left( {x + 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^4} - 5{x^3} + 10{x^2} - 10x + 4 < 0\\
\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) > 4x\left( {x + 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {{x^4} - 2{x^3}} \right) - \left( {3{x^3} - 6{x^2}} \right) + \left( {4{x^2} - 8x} \right) - \left( {2x - 4} \right) < 0\\
\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right) > 4x\left( {x + 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {{x^3} - 3{x^2} + 4x - 2} \right) < 0\\
\left( {x + 1} \right).\left[ {\left( {x - 1} \right)\left( {{x^2} + 1} \right) - 4x} \right] > 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 2} \right)\left[ {\left( {{x^3} - {x^2}} \right) - \left( {2{x^2} - 2x} \right) + \left( {2x - 2} \right)} \right] < 0\\
\left( {x + 1} \right).\left[ {{x^3} - {x^2} - 3x - 1} \right] > 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x - 1} \right)\left( {{x^2} - 2x + 2} \right) < 0\\
\left( {x + 1} \right).\left[ {\left( {{x^3} + {x^2}} \right) - \left( {2{x^2} + 2x} \right) - \left( {x + 1} \right)} \right] > 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x - 1} \right)\left( {{x^2} - 2x + 2} \right) < 0\\
\left( {x + 1} \right)\left( {x + 1} \right)\left( {{x^2} - 2x - 1} \right) > 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x - 1} \right)\left( {{x^2} - 2x + 2} \right) < 0\\
{\left( {x + 1} \right)^2}\left( {{x^2} - 2x - 1} \right) > 0
\end{array} \right.\,\,\,\,\,\,\left( 1 \right)\\
{x^2} - 2x + 2 = {\left( {x - 1} \right)^2} + 2 > 0,\,\;\;\forall x\\
{\left( {x + 1} \right)^2} \ge 0,\,\,\,\,\forall x\\
\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 1} \right)\left( {x - 2} \right) < 0\\
x \ne  - 1\\
{x^2} - 2x - 1 > 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
1 < x < 2\\
x \ne  - 1\\
\left[ \begin{array}{l}
x > 1 + \sqrt 2 \\
x < 1 - \sqrt 2 
\end{array} \right.
\end{array} \right.
\end{array}\) 

Ta thấy hệ trên vô nghiệm.

Vậy hệ phương trình đã cho vô nghiệm.