6.082. $\left\{\begin{array}{l}x^{2}+y^{4}=5 \\ x y^{2}=2\end{array}\right.$
## Solution.

From the second equation of the system $x=\frac{2}{y^{2}}$. Then from the first equation we get $\left(\frac{2}{y^{2}}\right)^{2}+y^{4}=5, y^{8}-5 y^{4}+4=0$. Hence, $y_{1}^{4}=1$ or $y_{2}^{4}=4$, from which $y_{1}=-1, y_{2}=1, y_{3}=-\sqrt{2}, y_{4}=\sqrt{2}$. Then $x_{1,2}=\frac{2}{1}=2$;

$$
x_{3,4}=\frac{2}{2}=1
$$

Answer: $(2 ; 1)(2 ;-1)(1 ; \sqrt{2})(1 ;-\sqrt{2})$