Three, (Total 25 points) Given a real number $t$ and a real number $k$ greater than $t$, let $a, b$ be positive numbers, and $\frac{1}{a+t}-\frac{1}{b+t}=\frac{1}{k}$. Determine the range of $b$ - $a$.
Three, from the given conditions we get
$b+t=\frac{1}{\frac{1}{a+t}-\frac{1}{k}}=\frac{k(a+t)}{k-a-t}$.
Therefore, from $a0$ we get
$k-t>x>0$.
From (2) we get $b-a=\frac{(k-x)^{2}}{x}$.
Let $\frac{(k-x)^{2}}{x}=y$, clearly $y>0$, then
$x^{2}-2 k x+k^{2}=y x$,

i.e., $x^{2}-(2 k+y) x+k^{2}=0$.
By the quadratic formula, we get
$x=\frac{2 k+y \pm \sqrt{(2 k+y)^{2}-4 k^{2}}}{2}$.
Since $x\frac{t^{2}}{k-t}$.
From (7), we get that the range of $b-a$ is any real number greater than $\frac{t^{2}}{k-t}$.
Note: For this problem, considering only the necessary conditions for $b-a$ is not sufficient; we must consider its necessary and sufficient conditions.
(Wu Weizhao, Henan Normal University, 453002)