6.192. $\left\{\begin{array}{l}\frac{4}{x+y}+\frac{4}{x-y}=3, \\ (x+y)^{2}+(x-y)^{2}=20 .\end{array}\right.$
## Solution.

Domain of definition: $x \neq \pm y$.

Let $\left\{\begin{array}{l}x+y=u, \\ x-y=v,\end{array}\right.$, where $u \neq 0$ and $v \neq 0$. Relative to $u$ and $v$, the system takes the form

$$
\left\{\begin{array} { l } 
{ \frac { 4 } { u } + \frac { 4 } { v } = 3 , } \\
{ u ^ { 2 } + v ^ { 2 } = 2 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
4(u+v)=3 u v, \\
(u+v)^{2}-2 u v=20 .
\end{array} \Rightarrow u v=\frac{4}{3}(u+v)\right.\right.
$$

From the second equation:

$$
\begin{aligned}
& (u+v)^{2}-2 \cdot \frac{4}{3}(u+v)=20 \Rightarrow 3(u+v)^{2}-8(u+v)-60=0, \\
& \text { from which }(u+v)_{1}=-\frac{10}{3} \text { or }(u+v)_{2}=6 ;(u v)_{1}=-\frac{40}{9},(u v)_{2}=\frac{4}{3} \cdot 6=8 ; \\
& \left\{\begin{array}{l}
u+v=-\frac{10}{3},\left\{\begin{array}{l}
u+v=6 \\
u v=8
\end{array}\right. \\
u v=-\frac{40}{9} ;
\end{array}\right.
\end{aligned}
$$

From here we find

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0355.jpg?height=222&width=840&top_left_y=652&top_left_x=101)

Then

$\left\{\begin{array}{l}x+y=\frac{-5-\sqrt{65}}{3}, \\ x-y=\frac{-5+\sqrt{65}}{3}\end{array} \Rightarrow\left\{\begin{array}{l}x_{1}=-\frac{5}{3}, \\ y_{1}=-\frac{\sqrt{65}}{3} .\end{array}\right.\right.$

$\left\{\begin{array}{l}x+y=\frac{-5+\sqrt{65}}{3}, \\ x-y=\frac{-5-\sqrt{65}}{3}\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}=-\frac{5}{3}, \\ y_{2}=\frac{\sqrt{65}}{3} .\end{array}\right.\right.$

$\left\{\begin{array}{l}x+y=2, \\ x-y=4\end{array} \Rightarrow\left\{\begin{array}{l}x_{3}=3 \\ y_{3}=-1\end{array}\right.\right.$

$\left\{\begin{array}{l}x+y=4, \\ x-y=2\end{array} \Rightarrow\left\{\begin{array}{l}x_{4}=3 \\ y_{4}=1\end{array}\right.\right.$

Answer: $\left(-\frac{5}{3} ;-\frac{\sqrt{65}}{3}\right),\left(-\frac{5}{3} ; \frac{\sqrt{65}}{3}\right),(3 ;-1),(3 ; 1)$.