One. (20 points) Given the equation in terms of $x$
$$
\frac{x^{2}+(5-2 m) x+m-3}{x-1}=2 x+m
$$

has no positive real roots. Find the range or value of $m$.


$$
\begin{array}{l}
x^{2}+(3 m-7) x+(3-2 m)=0 . \\
\text { By } \Delta=(3 m-7)^{2}-4(3-2 m) \\
=\left(3 m-\frac{17}{3}\right)^{2}+\frac{44}{9}>0,
\end{array}
$$

we know that equation (1) always has two distinct real roots. Let the two roots be $x_{1}$ and $x_{2}$. Then
$$
x_{1}+x_{2}=7-3 m, x_{1} x_{2}=3-2 m \text {. }
$$
(1) If equation (1) has two negative roots, then
$7-3 m<0 \Rightarrow m>\frac{7}{3}$ and $3-2 m>0 \Rightarrow m<\frac{3}{2}$. There is no solution for $m$.
(2) If 0 is a solution of equation (1), then
$$
3-2 m=0 \Rightarrow m=\frac{3}{2} \text {. }
$$

In this case, the solutions of equation (1) are $x=0, \frac{5}{2}$.
Since the original equation has no positive real roots, this is not consistent with the problem.
(3) If 1 is a solution of equation (1), then
$$
\begin{array}{l}
1+(3 m-7) \times 1+(3-2 m)=0 \\
\Rightarrow m=3 .
\end{array}
$$

In this case, the solutions of equation (1) are $x=1,-2$.
Upon verification, the solution of the original equation is -2, which meets the problem's requirements.
In summary, $m=3$.