One, (20 points) For what value of $a$ does the equation
$$
\frac{1}{\sqrt{(a-1) x^{2}+2 x-a}}=\frac{1}{\sqrt{2(a-1) x-2 a+10}}
$$

have exactly one root?
(1) When $a=1$, the original equation has a unique solution $x=\frac{9}{2}$.
(2) When $a \neq 1$, from equation (1) we get
$$
(a-1) x^{2}+2(2-a) x+a-10=0 \text {. }
$$

Its discriminant is
$$
\Delta=4(2-a)^{2}-4(a-1)(a-10)=4(7 a-6) \text {. }
$$

If $\Delta=0$, i.e., $a=\frac{6}{7}$, then $x=\frac{a-2}{a-1}=8$.
Substituting into equation (2) we get
$$
2(a-1) \frac{a-2}{a-1}-2 a+10=6>0 \text {. }
$$

Therefore, when $a=\frac{6}{7}$, the original equation has only one root $x=8$.
If $\Delta>0$, i.e., $a>\frac{6}{7}$, then the two roots of equation (3) are
$$
x_{1.2}=\frac{a-2 \pm \sqrt{7 a-6}}{a-1} \text {. }
$$

Substituting into the left side of equation (2) we get
$$
\begin{array}{l}
2(a-1) \frac{a-2 \pm \sqrt{7 a-6}}{a-1}-2 a+10 \\
=6 \pm 2 \sqrt{7 a-6} .
\end{array}
$$

To make the original equation have only one root, it must be
$$
\left\{\begin{array}{l}
6+2 \sqrt{7 a-6}>0, \\
6-2 \sqrt{7 a-6} \leqslant 0 .
\end{array}\right.
$$

When $a>\frac{6}{7}$, equation (4) holds.
It is only necessary to solve equation (5), getting $a \geqslant \frac{15}{7}$.
Therefore, when $a \geqslant \frac{15}{7}$, the original equation has only one root $x=6+2 \sqrt{7 a-6}$.

In summary, when $a=1$ or $a=\frac{6}{7}$ or $a \geqslant \frac{15}{7}$, the original equation has only one root.