Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
5{x^4} + {y^2} - 4{x^2}y - 20 = 0\\
\Leftrightarrow {x^4} + \left( {4{x^4} - 4{x^2}y + {y^2}} \right) = 20\\
\Leftrightarrow {x^4} + \left[ {{{\left( {2{x^2}} \right)}^2} - 2.2{x^2}.y + {y^2}} \right] = 20\\
\Leftrightarrow {x^4} + {\left( {2{x^2} - y} \right)^2} = 20\\
{\left( {2{x^2} - y} \right)^2} \ge 0,\,\,\,\,\forall x;y\\
\Rightarrow 0 \le {x^4} \le 20\\
x;y \in Z \Rightarrow \left[ \begin{array}{l}
{x^4} = 0 \Rightarrow {\left( {2{x^2} - y} \right)^2} = 20 \Leftrightarrow 2{x^2} - y = \pm 2\sqrt 5 \left( L \right)\\
{x^4} = 1 \Rightarrow {\left( {2{x^2} - y} \right)^2} = 19 \Rightarrow 2{x^2} - y = \pm \sqrt {19} \left( L \right)\\
{x^4} = {2^4} = 16 \Rightarrow {\left( {2{x^2} - y} \right)^2} = 4 \Leftrightarrow 2{x^2} - y = \pm 2\left( {t/m} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^4} = 16\\
2{x^2} - y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^4} = 16\\
2{x^2} - y = - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = \pm 2\\
y = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = \pm 2\\
y = 10
\end{array} \right.
\end{array} \right.
\end{array}\)
