Ta có:$\left\{\begin{matrix} \left | xy-4 \right |=8-y^{2} & \\ xy=2+x^{2} & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} \left | xy-4 \right |+y^{2}=8 & \\ 4xy=8+4x^{2} & \end{matrix}\right.$
$\Rightarrow 4x^{2}+8+y^{2}+\left | xy-4 \right |=4xy+8$
$\Leftrightarrow (2x-y)^{2}+\left | xy-4 \right |=0$
$\Leftrightarrow 2x=y;xy=4$
$\Leftrightarrow y=\pm 2\sqrt{2};x=\pm \sqrt{2}$.
When you have eliminated the impossible whatever remains, however improbable, must be the truth
__________SHERLOCK HOLMES____________