One. (20 points) Let $a, b$ be real numbers, and the equation with respect to $x$
$$
\frac{x}{x-1}+\frac{x-1}{x}=\frac{a+b x}{x^{2}-x}
$$

has no real roots. Find the value of the expression $8 a+4 b+|8 a+4 b-5|$.


One, the original equation can be transformed into
$$
2 x^{2}-(b+2) x+(1-a)=0 \text {. }
$$

Thus, $\Delta=(b+2)^{2}+8 a-8$.
(1) When $\Delta>0$, equation (1) has two distinct real roots.

Since the original equation has no solution, therefore, the two distinct real roots of equation (1) are 0 and 1, i.e.,
$$
\begin{aligned}
1-a & =0,2-(b+2)+(1-a)=0 \\
\Rightarrow a=1, b & =0 .
\end{aligned}
$$

Hence $8 a+4 b+|8 a+4 b-5|=11$.
In summary, the value of the algebraic expression is 5 or 11.