Let A = \ x ∈ R : |x^2 - 5x + 5| < 1\ and B = \ x ∈ R : (x - 2)/(x...

For set $A$, we have the inequality $|x^2 - 5x + 5| < 1$.
This is equivalent to $-1 < x^2 - 5x + 5 < 1$, which gives two inequalities:
1) $x^2 - 5x + 5 > -1 \Rightarrow x^2 - 5x + 6 > 0 \Rightarrow (x - 2)(x - 3) > 0$
$\Rightarrow x \in (-\infty, 2) \cup (3, \infty)$

2) $x^2 - 5x + 5 < 1 \Rightarrow x^2 - 5x + 4 < 0 \Rightarrow (x - 1)(x - 4) < 0$
$\Rightarrow x \in (1, 4)$

Taking the intersection of these two conditions, we get set $A$:
$A = (1, 2) \cup (3, 4)$

For set $B$, we have the rational inequality $\dfrac{x - 2}{x - 4} \le 0$.
The critical points are $x = 2$ and $x = 4$. Since the denominator cannot be zero, $x \neq 4$.
Using the wavy curve method, the solution is:
$B = [2, 4)$

Now, we find the intersection $A \cap B$:
$A \cap B = ((1, 2) \cup (3, 4)) \cap [2, 4)$
Notice that the interval $(1, 2)$ has no common elements with $[2, 4)$ since $2$ is not included in $(1, 2)$.
The interval $(3, 4)$ is fully contained within $[2, 4)$.
Thus, $A \cap B = (3, 4)$.

Answer: $(3, 4)$