One, (20 points) Given that $m$ is a real number, the function
$$
f(x)=m x^{2}+4 x-5-m
$$

has a root in the interval $[-2,2]$. Find the range of real values for $m$.
When $m=0$, the function has one root $\frac{5}{4}$.
When $m \neq 0$, the discriminant
$$
\Delta=16+4 m(m+5)=4(m+1)(m+4) \text{. }
$$
(1) When $\Delta=0$, $m=-1$ or -4.

If $m=-1$, the function has one root which is 2.
If $m=-4$, the function has one root which is $\frac{1}{2}$.
(2) When $\Delta>0$ and the function has only one root in $(-2,2)$, i.e., $f(-2) f(2)0, \\
(3 m-13)(3 m+3)0$ and the function has both roots within $(-2,2)$, then
$$
\left\{\begin{array}{l}
m > 0, \\
4 ( m + 1 ) ( m + 4 ) > 0, \\
- 2 < -\frac{m+5}{2} < 2, \\
3 m - 13 < 0, \\
3 m + 3 > 0
\end{array} \text{ or } \left\{\begin{array}{l}
m < 0, \\
-2 < -\frac{m+5}{2} < 2, \\
3 m - 13 > 0, \\
3 m + 3 < 0
\end{array}\right.
\right.
$$
This simplifies to
$$
\left\{\begin{array}{l}
m > 0, \\
m > -1, \\
-9 < m < -1, \\
m < \frac{13}{3}, \\
m > -1
\end{array} \text{ or } \left\{\begin{array}{l}
m < 0, \\
-9 < m < -1, \\
m > \frac{13}{3}, \\
m < -1
\end{array}\right.
\right.
$$
Thus, $m > 0$ and $m < -1$ or $m > \frac{13}{3}$ or $m \leqslant -4$.
(4) Only when $f(2) f(-2)=0$,
$$
m=-1 \text{ or } \frac{13}{3} \text{. }
$$

In summary, $m \geqslant -1$ or $m \geqslant -4$.