2. When $m$ is a negative real number, the equation about $x$
$$
x^{2}-2 x+1+m\left(\frac{1}{x}-2\right)=0
$$
(A) has three distinct real roots
(B) has two distinct real roots
(C) has one real root
(D) has no real roots
2.C.

The original equation can be transformed into $\frac{m}{x}=-(x-1)^{2}+2 m$. Let $y_{1}=\frac{m}{x}, y_{2}=-(x-1)^{2}+2 m$.

Draw the rough graphs of functions $y_{1}$ and $y_{2}$ (as shown in Figure 4). $y_{1}$ passes through the point $(1, m)$, and the vertex of $y_{2}$ is $(1,2 m)$. From the graph, it can be seen that the two function graphs have one intersection point. Therefore, the original equation has only one real root.