One. (20 points) The equation concerning $x$
$$
\sqrt{x^{2}-m}+2 \sqrt{x^{2}-1}=x
$$

has exactly one real root. Find the range of real values for $m$.
From equation (1), we know $x^{2} \geqslant m$, and $x \geqslant 1$.
If $mx \text {, }$
$$
$$

does not meet the requirements. Therefore, $m \geqslant 0$.
Transforming equation (1) yields
$$
\begin{array}{l}
2 \sqrt{x^{2}-1}=x-\sqrt{x^{2}-m} \\
\Rightarrow 2 x^{2}+m-4=-2 x \sqrt{x^{2}-m} \\
\Rightarrow 8(2-m) x^{2}=(m-4)^{2} .
\end{array}
$$

Thus, $0 \leqslant m<2$, and at this point, equation (1) can only have the solution
$$
x=\frac{4-m}{\sqrt{8(2-m)}} \text {. }
$$

Substituting $x=\frac{4-m}{\sqrt{8(2-m)}}$ into equation (1), and simplifying yields $|3 m-4|=4-3 m \Rightarrow 0 \leqslant m \leqslant \frac{4}{3}$.
At this point, equation (1) has a unique solution $x=\frac{4-m}{\sqrt{8(2-m)}}$. In summary, the range of real numbers $m$ is $0 \leqslant m \leqslant \frac{4}{3}$.