If `(x^(2)-10x+13)/((x-1)(x^(2)-5x+6))=A/(x-1)+B/(x-2)+C/(x-3)`, then C =

Options

A. -1
B. -2
C. -3
D. -4

Answer

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 need to find the value of \(C\).

### Step 1: Factor the denominator
The denominator can be factored as follows:

\[ x^2 - 5x + 6 = (x-2)(x-3). \]

Thus, 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 2: Combine the right-hand side
To combine the right-hand side, we need a common denominator:

\[ \frac{A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)}{(x-1)(x-2)(x-3)}. \]

### Step 3: 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 4: 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 5: Set up equations from coefficients
Now, we can set up a system of equations based on the coefficients:

1. \(A + B + C = 1\) (coefficient of \(x^2\))
2. \(-5A - 4B - 3C = -10\) (coefficient of \(x\))
3. \(6A + 3B + 2C = 13\) (constant term)

### Step 6: Solve the system of equations
From equation (1):

\[ C = 1 - A - B. \]

Substituting \(C\) into equations (2) and (3):

1. \(-5A - 4B - 3(1 - A - B) = -10\)
\[ -5A - 4B - 3 + 3A + 3B = -10 \implies -2A - B = -7 \implies 2A + B = 7 \quad \text{(Equation 4)} \]

2. \(6A + 3B + 2(1 - A - B) = 13\)
\[ 6A + 3B + 2 - 2A - 2B = 13 \implies 4A + B = 11 \quad \text{(Equation 5)} \]

Now we have:

- Equation 4: \(2A + B = 7\)
- Equation 5: \(4A + B = 11\)

### Step 7: Subtract the equations
Subtract Equation 4 from Equation 5:

\[ (4A + B) - (2A + B) = 11 - 7 \implies 2A = 4 \implies A = 2. \]

### Step 8: Substitute back to find \(B\)
Substituting \(A = 2\) into Equation 4:

\[ 2(2) + B = 7 \implies 4 + B = 7 \implies B = 3. \]

### Step 9: Find \(C\)
Now substitute \(A\) and \(B\) back into the expression for \(C\):

\[ C = 1 - A - B = 1 - 2 - 3 = -4. \]

### Final Answer
Thus, the value of \(C\) is

\[ \boxed{-4}. \]