$\lim_{x \to 1} \left[\dfrac{x-2}{x^2-x} - \dfrac{1}{x^3 - 3x^2 + 2x}\right] = $

The given limit is $\lim_{x \to 1} \left[\dfrac{x-2}{x^2-x} - \dfrac{1}{x^3 - 3x^2 + 2x}\right]$

Factorizing the denominators, we get:

$= \lim_{x \to 1} \left[\dfrac{x-2}{x(x-1)} - \dfrac{1}{x(x-1)(x-2)}\right]$

Taking the common denominator $x(x-1)(x-2)$:

$= \lim_{x \to 1} \left[\dfrac{(x-2)^2 - 1}{x(x-1)(x-2)}\right]$

Expanding the numerator:

$= \lim_{x \to 1} \left[\dfrac{x^2 - 4x + 4 - 1}{x(x-1)(x-2)}\right]$

$= \lim_{x \to 1} \left[\dfrac{x^2 - 4x + 3}{x(x-1)(x-2)}\right]$

Factorizing the numerator:

$= \lim_{x \to 1} \left[\dfrac{(x-1)(x-3)}{x(x-1)(x-2)}\right]$

Canceling the common factor $(x-1)$ since $x \to 1$ implies $x \neq 1$:

$= \lim_{x \to 1} \dfrac{x-3}{x(x-2)}$

Substituting $x = 1$ into the simplified expression:

$= \dfrac{1-3}{1(1-2)} = \dfrac{-2}{-1} = 2$

Answer: $2$