If `(x^(2)+1)/((x^(2)+2)(x^(2)+3))=(A)/(x^(2)+3)+(B)/(x^(2)+2)` then `(A, B)=`

Options

A. (1, 2)
B. (2, -1)
C. (1, -2)
D. (-2, -1)

Answer

To solve the equation

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

we will find the values of \(A\) and \(B\).

### Step 1: Combine the right-hand side into a single fraction

We start by finding a common denominator for the right-hand side:

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

### Step 2: Set the numerators equal

Now we can set the numerators equal to each other since the denominators are the same:

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

### Step 3: Expand the right-hand side

Expanding the right-hand side gives:

\[ x^2 + 1 = Ax^2 + 2A + Bx^2 + 3B \]

Combining like terms, we have:

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

### Step 4: Compare coefficients

Now we can compare coefficients from both sides of the equation.

1. Coefficient of \(x^2\):
\[ 1 = A + B \quad \text{(1)} \]

2. Constant term:
\[ 1 = 2A + 3B \quad \text{(2)} \]

### Step 5: Solve the system of equations

Now we have a system of two equations:

1. \(A + B = 1\)
2. \(2A + 3B = 1\)

From equation (1), we can express \(A\) in terms of \(B\):

\[ A = 1 - B \]

Substituting \(A\) into equation (2):

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

Expanding this gives:

\[ 2 - 2B + 3B = 1 \]

Combining like terms results in:

\[ 2 + B = 1 \]

### Step 6: Solve for \(B\)

Now, we can isolate \(B\):

\[ B = 1 - 2 = -1 \]

### Step 7: Substitute back to find \(A\)

Now substituting \(B = -1\) back into equation (1):

\[ A + (-1) = 1 \implies A = 1 + 1 = 2 \]

### Final Result

Thus, the values of \(A\) and \(B\) are:

\[ (A, B) = (2, -1) \]