`lim_(xrarr1) [(x-2)/(x^(2)-x)-(1)/(x^(3)-3x^(2)+2x)]`

Answer

To solve the limit \( \lim_{x \to 1} \left( \frac{x-2}{x^2 - x} - \frac{1}{x^3 - 3x^2 + 2x} \right) \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{x-2}{x^2 - x} - \frac{1}{x^3 - 3x^2 + 2x} \]

### Step 2: Factor the denominators
The denominator \( x^2 - x \) can be factored as:
\[ x^2 - x = x(x-1) \]
The denominator \( x^3 - 3x^2 + 2x \) can be factored as:
\[ x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x-1)(x-2) \]

### Step 3: Rewrite the limit with factored denominators
Now we can rewrite the limit:
\[ \lim_{x \to 1} \left( \frac{x-2}{x(x-1)} - \frac{1}{x(x-1)(x-2)} \right) \]

### Step 4: Find a common denominator
The common denominator for the two fractions is \( x(x-1)(x-2) \). Rewriting the limit gives:
\[ \lim_{x \to 1} \left( \frac{(x-2)(x-2) - 1}{x(x-1)(x-2)} \right) \]
This simplifies to:
\[ \lim_{x \to 1} \frac{(x-2)^2 - 1}{x(x-1)(x-2)} \]

### Step 5: Simplify the numerator
The numerator can be simplified:
\[ (x-2)^2 - 1 = (x-2-1)(x-2+1) = (x-3)(x-1) \]
Thus, we have:
\[ \lim_{x \to 1} \frac{(x-3)(x-1)}{x(x-1)(x-2)} \]

### Step 6: Cancel common factors
We can cancel \( (x-1) \) from the numerator and denominator:
\[ \lim_{x \to 1} \frac{x-3}{x(x-2)} \]

### Step 7: Substitute \( x = 1 \)
Now we substitute \( x = 1 \):
\[ \frac{1-3}{1(1-2)} = \frac{-2}{1 \cdot -1} = 2 \]

### Final Answer
Thus, the limit is:
\[ \boxed{2} \]