Đáp án:
$\begin{array}{l}
Dxkd:x \ne 0;y \ne 0\\
{x^2} + {y^2} + \frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} = 4\\
\Rightarrow \left( {{x^2} + 2 + \frac{1}{{{x^2}}}} \right) + \left( {{y^2} + 2 + \frac{1}{{{y^2}}}} \right) = 8\\
\Rightarrow {\left( {x + \frac{1}{x}} \right)^2} + {\left( {y + \frac{1}{y}} \right)^2} = 8\\
Đặt:\left\{ \begin{array}{l}
x + \frac{1}{x} = a\\
y + \frac{1}{y} = b
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a + b = 4\\
{a^2} + {b^2} = 8
\end{array} \right.\\
\Rightarrow a = b = 2\\
\Rightarrow x + \frac{1}{x} = y + \frac{1}{y} = 2\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + 1 = 2x\\
{y^2} + 1 = 2y
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x^2} - 2x + 1 = 0\\
{y^2} - 2y + 1 = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 1\\
y = 1
\end{array} \right.
\end{array}$
Vậy x=y=1
