To solve the equation
\[
\frac{x^2 - 10x + 13}{(x-1)(x^2 - 5x + 6)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3},
\]
we will first factor the denominator on the left-hand side.
### Step 1: Factor the denominator
The quadratic \(x^2 - 5x + 6\) can be factored as \((x-2)(x-3)\). Thus, the complete factorization of the denominator is:
\[
(x-1)(x-2)(x-3).
\]
### Step 2: Rewrite the equation
Now we can rewrite the equation as:
\[
\frac{x^2 - 10x + 13}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}.
\]
### Step 3: Combine the right-hand side
To combine the right-hand side, we need a common denominator, which is \((x-1)(x-2)(x-3)\). Thus, we can rewrite it as:
\[
\frac{A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)}{(x-1)(x-2)(x-3)}.
\]
### Step 4: Set the numerators equal
Now we equate the numerators:
\[
x^2 - 10x + 13 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2).
\]
### Step 5: Expand the right-hand side
Expanding each term:
1. \(A(x-2)(x-3) = A(x^2 - 5x + 6)\)
2. \(B(x-1)(x-3) = B(x^2 - 4x + 3)\)
3. \(C(x-1)(x-2) = C(x^2 - 3x + 2)\)
Combining these gives:
\[
(A + B + C)x^2 + (-5A - 4B - 3C)x + (6A + 3B + 2C).
\]
### Step 6: Compare coefficients
Now we compare coefficients from both sides:
1. For \(x^2\): \(A + B + C = 1\) (Equation 1)
2. For \(x\): \(-5A - 4B - 3C = -10\) (Equation 2)
3. For the constant term: \(6A + 3B + 2C = 13\) (Equation 3)
### Step 7: Solve the system of equations
From Equation 1, we can express \(C\):
\[
C = 1 - A - B.
\]
Substituting \(C\) into Equations 2 and 3:
**Substituting into Equation 2:**
\[
-5A - 4B - 3(1 - A - B) = -10,
\]
\[
-5A - 4B - 3 + 3A + 3B = -10,
\]
\[
-2A - B = -7 \quad \Rightarrow \quad 2A + B = 7. \quad \text{(Equation 4)}
\]
**Substituting into Equation 3:**
\[
6A + 3B + 2(1 - A - B) = 13,
\]
\[
6A + 3B + 2 - 2A - 2B = 13,
\]
\[
4A + B - 11 = 0 \quad \Rightarrow \quad 4A + B = 11. \quad \text{(Equation 5)}
\]
### Step 8: Solve Equations 4 and 5
Now we solve Equations 4 and 5:
1. From Equation 4: \(B = 7 - 2A\).
2. Substitute into Equation 5:
\[
4A + (7 - 2A) = 11,
\]
\[
2A + 7 = 11,
\]
\[
2A = 4 \quad \Rightarrow \quad A = 2.
\]
### Step 9: Find B and C
Substituting \(A = 2\) back into Equation 4:
\[
2(2) + B = 7 \quad \Rightarrow \quad 4 + B = 7 \quad \Rightarrow \quad B = 3.
\]
Now substitute \(A\) and \(B\) into Equation 1 to find \(C\):
\[
2 + 3 + C = 1 \quad \Rightarrow \quad C = 1 - 5 = -4.
\]
### Step 10: Final values
Thus, we have:
- \(A = 2\)
- \(B = 3\)
- \(C = -4\)
### Step 11: Arrange in ascending order
Now we arrange \(A\), \(B\), and \(C\) in ascending order:
\[
C < A < B \quad \Rightarrow \quad -4 < 2 < 3.
\]
So the final answer in ascending order is:
\[
C, A, B \quad \Rightarrow \quad -4, 2, 3.
\]
