If `(x^(2)-5x+7)/(x-1)^(3)=A/(x-1)+B/(x-1)^(2)+C/(x-1)^(3) " then " A+B-C=`

Options

A. 5
B. 3
C. 0
D. -5

Answer

To solve the equation

\[ \frac{x^2 - 5x + 7}{(x - 1)^3} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} \]

we will follow these steps:

### Step 1: Combine the Right Side
First, we need to combine the right-hand side into a single fraction. The common denominator is \((x - 1)^3\):

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

### Step 2: Set the Numerators Equal
Now that we have a common denominator, we can set the numerators equal to each other:

\[ x^2 - 5x + 7 = A(x - 1)^2 + B(x - 1) + C \]

### Step 3: Expand the Right Side
Next, we expand the right-hand side:

\[ A(x - 1)^2 = A(x^2 - 2x + 1) = Ax^2 - 2Ax + A \]
\[ B(x - 1) = Bx - B \]

Combining these gives:

\[ Ax^2 - 2Ax + A + Bx - B + C = Ax^2 + (-2A + B)x + (A - B + C) \]

### Step 4: Compare Coefficients
Now we equate the coefficients from both sides of the equation:

1. Coefficient of \(x^2\):
\[ A = 1 \]

2. Coefficient of \(x\):
\[ -2A + B = -5 \]

3. Constant term:
\[ A - B + C = 7 \]

### Step 5: Solve for A, B, and C
From \(A = 1\), we substitute into the second equation:

\[ -2(1) + B = -5 \implies -2 + B = -5 \implies B = -3 \]

Now substitute \(A\) and \(B\) into the third equation:

\[ 1 - (-3) + C = 7 \implies 1 + 3 + C = 7 \implies C = 7 - 4 = 3 \]

### Step 6: Calculate A + B - C
Now we have:
- \(A = 1\)
- \(B = -3\)
- \(C = 3\)

We need to find \(A + B - C\):

\[ A + B - C = 1 - 3 - 3 = 1 - 6 = -5 \]

### Final Answer
Thus, the value of \(A + B - C\) is:

\[ \boxed{-5} \]