`x^(3)/((2x-1)(x+2)(x-3))=A+B/(2x-1)+C/(x+2)+D/(x-3) rArr A =`

Options

A. `1/2`
B. `- 1/50`
C. `- 8/25`
D. `- 27/25`

Answer

To solve the equation

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

we need to find the value of \( A \).

### Step 1: Set Up the Equation
We start by expressing the left-hand side in terms of the right-hand side. We can write:

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

### Step 2: Expand the Right-Hand Side
Next, we will expand the right-hand side. This involves multiplying out the terms:

1. **Expand \( A(2x-1)(x+2)(x-3) \)**:
- First, expand \( (x+2)(x-3) = x^2 - x - 6 \).
- Then, expand \( (2x-1)(x^2 - x - 6) = 2x^3 - 2x^2 - 12x - x^2 + x + 6 = 2x^3 - 3x^2 - 11x + 6 \).
- Finally, multiply by \( A \): \( A(2x^3 - 3x^2 - 11x + 6) \).

2. **Expand \( B(x+2)(x-3) \)**:
- \( B(x^2 - x - 6) \).

3. **Expand \( C(2x-1)(x-3) \)**:
- \( C(2x^2 - 6x - x + 3) = C(2x^2 - 7x + 3) \).

4. **Expand \( D(2x-1)(x+2) \)**:
- \( D(2x^2 + 4x - x - 2) = D(2x^2 + 3x - 2) \).

### Step 3: Combine All Terms
Now we combine all the expanded terms:

\[ x^3 = A(2x^3 - 3x^2 - 11x + 6) + B(x^2 - x - 6) + C(2x^2 - 7x + 3) + D(2x^2 + 3x - 2) \]

### Step 4: Collect Like Terms
Group the coefficients of \( x^3 \), \( x^2 \), \( x \), and the constant term:

- Coefficient of \( x^3 \): \( 2A \)
- Coefficient of \( x^2 \): \( -3A + B + 2C + 2D \)
- Coefficient of \( x \): \( -11A - B - 7C + 3D \)
- Constant term: \( 6A - 6B + 3C - 2D \)

### Step 5: Set Up the System of Equations
From the equation \( x^3 = 0x^2 + 0x + 0 \), we can set up the following equations:

1. \( 2A = 1 \) (from the coefficient of \( x^3 \))
2. \( -3A + B + 2C + 2D = 0 \) (from the coefficient of \( x^2 \))
3. \( -11A - B - 7C + 3D = 0 \) (from the coefficient of \( x \))
4. \( 6A - 6B + 3C - 2D = 0 \) (from the constant term)

### Step 6: Solve for \( A \)
From the first equation, we can solve for \( A \):

\[ A = \frac{1}{2} \]

### Step 7: Conclusion
Thus, the value of \( A \) is

\[ \boxed{\frac{1}{2}} \]