5. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}y-2 \sqrt{x y}-\sqrt{\frac{y}{x}}+2=0, \\ 3 x^{2} y^{2}+y^{4}=84 .\end{array}\right.$ Answer: $\left(\frac{1}{3} ; 3\right),\left(\sqrt[4]{\frac{21}{76}} ; 2 \cdot \sqrt[4]{\frac{84}{19}}\right)$.
Solution. Let $\sqrt{\frac{y}{x}}=u, \sqrt{x y}=v \quad$ (with $u>0, \quad v>0$). Then $\quad u v=\sqrt{\frac{y}{x}} \cdot \sqrt{x y}=\sqrt{y^{2}}=|y|=y$, $\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{y}{x}}=\sqrt{x^{2}}=|x|=x$, since by the condition $x$ and $y$ are positive. The system takes the form

$$
\left\{\begin{array}{l}
u v - 2 v - u + 2 = 0, \\
3 v^{4} + u^{4} v^{4} = 84
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(v-1)(u-2)=0 \\
3 v^{4} + u^{4} v^{4} = 84
\end{array}\right.\right.
$$

From the first equation, it follows that $v=1$ or $u=2$.

If $v=1$, then $3+u^{4}=84$, from which $u=3$; then $x=\frac{v}{u}=\frac{1}{3}, y=u v=3$.

If $u=2$, then $3 v^{4}+16 v^{4}=84$, from which $v=4 \sqrt{\frac{84}{19}}$; then $x=\frac{v}{u}=\sqrt[4]{\frac{21}{76}}, y=u v=2 \cdot 4 \sqrt{\frac{84}{19}}$.