One. (20 points) Let $a$ be an integer such that the equation $a x^{2}-(a+5) x+a+7=0$ has at least one rational root. Find all possible rational roots of the equation.


When $a=0$, the rational root of the equation is $x=\frac{7}{5}$.
The following considers the case where $a \neq 0$. In this case, the original equation is a quadratic equation, and by the discriminant
$$
(a+5)^{2}-4 a(a+7) \geqslant 0,
$$

which simplifies to $3 a^{2}+18 a-25 \leqslant 0$.
Solving this, we get $-\frac{-9-\sqrt{156}}{3} \leqslant a \leqslant \frac{-9+\sqrt{156}}{3}$.
The integer $a$ can only take non-zero integer values from $1, -1$, $-2, -3, -4, -5, -6, -7$. From the equation, we have
$$
x=\frac{a+5 \pm \sqrt{52-3(a+3)^{2}}}{2 a} \text{. }
$$

When $a=1$, from equation (1) we get $x=2$ and 4;
When $a=-1$, the equation has no rational roots;
When $a=-2$, from equation (1) we get $x=1$ and $-\frac{5}{2}$;
When $a=-3$, the equation has no rational roots;
When $a=-4$, from equation (1) we get $x=-1$ and $\frac{3}{4}$;
When $a=-5$, the equation has no rational roots;
When $a=-6$, from equation (1) we get $x=\frac{1}{2}$ and $-\frac{1}{3}$;
When $a=-7$, from equation (1) we get $x=0$ and $\frac{2}{7}$.
Therefore, for different values of $a$, the equation has a total of 11 rational roots.