: Let $x>1, y>1$, the equation $x-y+\frac{3}{x-1}+\frac{3}{y-1}=2(\sqrt{x+2}+\sqrt{y+2})$ has the solution $(\mu, y)=$
12. $\left(\frac{3+\sqrt{13}}{2}, \frac{3+\sqrt{13}}{2}\right)$.

Since $x+\frac{3}{x-1}=x-1+\frac{x+2}{x-1} \geqslant 2 \sqrt{x+2}$, similarly $y+\frac{3}{y-1} \geqslant 2 \sqrt{y+2}$, now the equality holds, so $x-1=\frac{x+2}{x-1}, y-1=\frac{y+2}{y-1}$. Solving these, we get $x=y=\frac{3+\sqrt{13}}{2}$.