6.183. $\left\{\begin{array}{l}x y(x+1)(y+1)=72, \\ (x-1)(y-1)=2 .\end{array}\right.$
## Solution.

From the condition $\left\{\begin{array}{l}x y(x y+x+y+1)=72, \\ x y-(x+y)=1 .\end{array}\right.$

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

$\left\{\begin{array}{l}u(u+v+1)=72, \\ u-v=1\end{array} \Rightarrow v=u-1\right.$. Substituting this value of $v$ into the first equation of the system, we get $u(u+u-1+1)=72,2 u^{2}=72$, from which $u_{1}=-6$, $u_{2}=6$ and $v_{1}=-7, v_{2}=5$.

Then the given system is equivalent to the combination of two systems:

$\left\{\begin{array}{l}x y=-6, \\ x+y=-7\end{array}\right.$ or

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0348.jpg?height=220&width=1054&top_left_y=419&top_left_x=143)

Answer: $(2 ; 3),(3 ; 2),\left(\frac{-7+\sqrt{73}}{2} ;-\frac{7+\sqrt{73}}{2}\right),\left(-\frac{7+\sqrt{73}}{2} ; \frac{-7+\sqrt{73}}{2}\right)$.