6.240. $\left\{\begin{array}{l}\sqrt{\frac{x}{y}+2+\frac{y}{x}}=\frac{5}{2}, \\ |x+y|=5 .\end{array}\right.$
Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0 .\end{array}\right.$

Transform the first equation of the system

$\sqrt{\frac{x^{2}+2 x y+y^{2}}{x y}}=\frac{5}{2} \Leftrightarrow \frac{\sqrt{(x+y)^{2}}}{\sqrt{x y}}=\frac{5}{2} \Leftrightarrow \frac{|x+y|}{\sqrt{x y}}=\frac{5}{2}$.

Thus, we have:

$\left\{\begin{array}{l}\frac{|x+y|}{\sqrt{x y}}=\frac{5}{2} \\ |x+y|=5\end{array}, \Leftrightarrow\left\{\begin{array}{l}\frac{5}{\sqrt{x y}}=\frac{5}{2}, \\ |x+y|=5\end{array}, \Leftrightarrow\left\{\begin{array}{l}\sqrt{x y}=2, \\ |x+y|=5\end{array} \Rightarrow\right.\right.\right.$

$\Rightarrow\left\{\begin{array}{l}x y=4, \\ |x+y|=5 .\end{array}\right.$

We get two cases:

1) $\left\{\begin{array}{l}x y=4, \\ x+y=-5 ;\end{array}\right.$, 2) $\left\{\begin{array}{l}x y=4, \\ x+y=5 .\end{array}\right.$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0388.jpg?height=117&width=1068&top_left_y=914&top_left_x=144)

Answer: $(-4 ;-1),(-1 ;-4),(1 ; 4),(4 ; 1)$.