One, (14 points) Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+\frac{9}{x}+\frac{4}{y}=10, \\
\left(x^{2}+9\right)\left(y^{2}+4\right)=24 x y .
\end{array}\right.
$$
$$
\left\{\begin{array}{l}
x+\frac{9}{x}+y+\frac{4}{y}=10, \\
\left(x+\frac{9}{x}\right)\left(y+\frac{4}{y}\right)=24 .
\end{array}\right.
$$

Thus, $x+\frac{9}{x}$ and $y+\frac{4}{y}$ can be seen as the two roots of the equation $t^{2}-10 t+24=0$.
By $(t-4)(t-6)=0$, we get $t_{1}=4, t_{2}=6$.

We have (I) $\left\{\begin{array}{l}x+\frac{9}{x}=4, \\ y+\frac{4}{y}=6\end{array}\right.$ or (II) $\left\{\begin{array}{l}x+\frac{9}{x}=6, \\ y+\frac{4}{y}=4 .\end{array}\right.$
From equation set (I), we get $\left\{\begin{array}{l}x^{2}-4 x+9=0, \\ y^{2}-6 y+4=0 .\end{array}\right.$ (no solution)
From equation set (II), we get $\left\{\begin{array}{l}x^{2}-6 x+9=0, \\ y^{2}-4 y+4=0 .\end{array}\right.$
That is $\left\{\begin{array}{l}(x-3)^{2}=0, \\ (y-2)^{2}=0 .\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}x=3, \\ y=2 .\end{array}\right.$. After checking, it fits.