If `(x)/((1+x^(2))(3-2x))=(A)/(3-2x)+(Bx+C)/(1+x^(2))` then `C=`

Options

A. `(-1)/(13)`
B. `(2)/(13)`
C. `(-2)/(13)`
D. `(1)/(13)`

Answer

To solve the equation

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

we will find the value of \( C \).

### Step 1: Combine the Right Side
We start by combining the fractions on the right side:

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

### Step 2: Set the Numerators Equal
Now, we equate the numerators since the denominators are the same:

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

### Step 3: Expand the Right Side
Expanding the right side gives:

\[ x = A + Ax^2 + (3Bx + 3C - 2Bx^2 - 2Cx) \]

Combining like terms:

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

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

1. Coefficient of \( x^2 \):
\[ 0 = A - 2B \quad \text{(1)} \]

2. Coefficient of \( x \):
\[ 1 = 3B - 2C \quad \text{(2)} \]

3. Constant term:
\[ 0 = A + 3C \quad \text{(3)} \]

### Step 5: Solve the System of Equations
From equation (1), we can express \( A \) in terms of \( B \):
\[ A = 2B \quad \text{(4)} \]

Substituting (4) into equation (3):
\[ 0 = 2B + 3C \implies 3C = -2B \implies C = -\frac{2}{3}B \quad \text{(5)} \]

Now substitute (5) into equation (2):
\[ 1 = 3B - 2\left(-\frac{2}{3}B\right) \]
\[ 1 = 3B + \frac{4}{3}B \]
\[ 1 = \frac{9B + 4B}{3} = \frac{13B}{3} \]
\[ B = \frac{3}{13} \]

### Step 6: Find \( C \)
Now substitute \( B \) back into equation (5):
\[ C = -\frac{2}{3} \cdot \frac{3}{13} = -\frac{2}{13} \]

Thus, the value of \( C \) is

\[ \boxed{-\frac{2}{13}} \]