One, (20 points) $k$ is the minimum value of $\sqrt{\frac{5 m^{2}-2 \sqrt{5} m+7}{m^{2}+1}}$ as $m$ takes all real values. Solve the system of equations for $x, y$ in the real numbers:
$$
\left\{\begin{array}{l}
\sqrt{x^{2}-\frac{6}{x}}+\sqrt{y^{2}-y}=1+k, \\
y\left(x^{2} y-x^{2}+\frac{6}{x}-\frac{6 y}{x}\right)=\sqrt{6}\left(5-k^{2}\right) .
\end{array}\right.
$$
Let $p=\frac{5 m^{2}-2 \sqrt{5} m+7}{m^{2}+1}$. Then
$$
\begin{array}{l}
(p-5) m^{2}+2 \sqrt{5} m+p-7=0 . \\
\Delta \geqslant 0 \Rightarrow p^{2}-12 p+30 \leqslant 0,
\end{array}
$$

Thus, $6-\sqrt{6} \leqslant p \leqslant 6+\sqrt{6}$.
When $m=\frac{\sqrt{5}}{\sqrt{6}-1}$, $p=6-\sqrt{6}$,
then $k=\sqrt{p_{\text {min }}}=\sqrt{6-\sqrt{6}}$.
Substituting $k=\sqrt{6-\sqrt{6}}$ into the system of equations and noting that $x^{2}-\frac{6}{x} \geqslant 0, y^{2}-y \geqslant 0$, the system of equations can be rewritten as
$$
\left\{\begin{aligned}
\sqrt{x^{2}-\frac{6}{x}}+\sqrt{y^{2}-y} & =1+\sqrt{6-\sqrt{6}}, \\
\sqrt{x^{2}-\frac{6}{x}} \cdot \sqrt{y^{2}-y} & =\sqrt{y\left(x^{2} y-x^{2}+\frac{6}{x}-\frac{6 y}{x}\right)} \\
& =1 \times \sqrt{6-\sqrt{6}} .
\end{aligned}\right.
$$

Thus, $\sqrt{x^{2}-\frac{6}{x}}$ and $\sqrt{y^{2}-y}$ are the roots of the equation
$$
t^{2}-(1+\sqrt{6-\sqrt{6}}) t+\sqrt{6-\sqrt{6}}=0
$$

(1). $\sqrt{x^{2}-\frac{6}{x}}=1, \sqrt{y^{2}-y}=\sqrt{6-\sqrt{6}}$;
(2). $\sqrt{x^{2}-\frac{6}{x}}=\sqrt{6-\sqrt{6}}, \sqrt{y^{2}-y}=1$.
For (1):
$$
\left\{\begin{array}{l}\frac{(x-2)\left(x^{2}+2 x+3\right)}{x}=0 \Rightarrow x=2 . \\ y^{2}-y=6-\sqrt{6} \Rightarrow y=\sqrt{6} \text { or } y=1-\sqrt{6} .\end{array}\right.
$$
For (2):
$$
\left\{\begin{array}{l}\frac{(x-\sqrt{6})\left(x^{2}+\sqrt{6} x+\sqrt{6}\right)}{x}=0 \Rightarrow x=\sqrt{6} . \\ y^{2}-y=1 \Rightarrow y=\frac{1 \pm \sqrt{5}}{2} .\end{array}\right.
$$
Upon verification, there are no extraneous roots. Therefore, the solutions to the system of equations are