6.241. $\left\{\begin{array}{l}u-v+\sqrt{\frac{u-v}{u+v}}=\frac{12}{u+v} \\ u^{2}+v^{2}=41 .\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}\frac{u-v}{u+v} \geq 0, \\ u+v \neq 0 .\end{array}\right.$

This system is equivalent to two systems of equations:

1) $\left\{\begin{array}{l}u-v \geq 0, \\ u+v>0, \\ u-v+\sqrt{\frac{u-v}{u+v}}=\frac{12}{u+v}, \\ u^{2}+v^{2}=41\end{array}\right.$
2) $\left\{\begin{array}{l}u-v \leq 0, \\ u+v<0, \\ u-v+\sqrt{\frac{u-v}{u+v}}=\frac{12}{u+v}, \\ u^{2}+v^{2}=41\end{array}\right.$

For system 1), multiplying the first equation by $u+v>0$, we get $u^{2}-v^{2}+\sqrt{u^{2}-v^{2}}-12=0 \Rightarrow \sqrt{u^{2}-v^{2}}=3, u^{2}-v^{2}=9$ or $\sqrt{u^{2}-v^{2}}=-4$ (no solutions). Thus, system 1) becomes $\left\{\begin{array}{l}u^{2}-v^{2}=9, \\ u^{2}+v^{2}=41\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}u^{2}=25, \\ v^{2}=16\end{array} \Rightarrow\left\{\begin{array}{l}u= \pm 5 \\ v= \pm 4\end{array}\right.\right.$

We obtain:

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0389.jpg?height=127&width=558&top_left_y=889&top_left_x=155)

Since $u-v \geq 0$ and $u+v>0$, we have $\left\{\begin{array}{l}u_{1}=5, \\ v_{1}=-4 ;\end{array}\left\{\begin{array}{l}u_{2}=5, \\ v_{2}=4 .\end{array}\right.\right.$

For system 2), multiplying the first equation by $u+v<0$, we get

$$
u^{2}-v^{2}-\sqrt{u^{2}-v^{2}}-12=0 \Rightarrow \sqrt{u^{2}-v^{2}}=4, u^{2}-v^{2}=16
$$

or $\sqrt{u^{2}-v^{2}}=-3$ (no solutions).

System 2) becomes $\left\{\begin{array}{l}u^{2}-v^{2}=16, \\ u^{2}+v^{2}=41 .\end{array}\right.$

From here, considering $u-v \leq 0$ and $u+v<0$,

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0389.jpg?height=127&width=459&top_left_y=1616&top_left_x=154)

Answer: $(5 ;-4),(5 ; 4),(-\sqrt{28.5} ;-\sqrt{12.5}),(-\sqrt{28.5} ; \sqrt{12.5})$.