Solve the following equation:

$$
|1-\sqrt{x-2}|\left(-x^{2}+4 x-3\right)=1
$$
$$
|1-\sqrt{x-2}|\left(-x^{2}+4 x-3\right)=1
$$

On the left side of equation (1), $\sqrt{x-2}$ appears, so only $x$ values for which $x \geq 2$ are considered. Furthermore, the first factor on the left side of (1) is non-negative, so the second factor must be positive; otherwise, their product could not be 1:

$$
0<-x^{2}+4 x-3=(3-x)(x-1)
$$

From this, $1<x<3$, or, in conjunction with the previous constraint,

$$
2 \leq x<3
$$

Therefore, it is sufficient to look for the solutions of (1) among the $x$ values that satisfy (2).

If (2) holds for $x$, then

$$
0<1-\sqrt{x-2} \leq 1
$$

(and thus the absolute value signs can be omitted),

$$
0<-x^{2}+4 x-3=1-(x-2)^{2} \leq 1
$$

Moreover, according to (1), $1-\sqrt{x-2}$ and $-x^{2}+4 x-3$ are reciprocals of each other; according to (3) and (4), both values fall between 0 and 1. This is only possible if both values are 1:

$$
1-\sqrt{x-2}=-x^{2}+4 x-3=1
$$

From this, $x=2$. Since this also satisfies the inequality (2), $x=2$ is the only solution to the equation.