If `(ax-1)/((1-x+x^(2))(2+x))=(x)/(1-x+x^(2))-(1)/(2+x)` then `a=`

Options

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

Answer

To solve the equation

\[ \frac{ax-1}{(1-x+x^2)(2+x)} = \frac{x}{1-x+x^2} - \frac{1}{2+x} \]

we will follow these steps:

### Step 1: Multiply both sides by the common denominator
Multiply both sides of the equation by \((1-x+x^2)(2+x)\) to eliminate the denominators.

\[ (ax - 1) = \left( \frac{x}{1-x+x^2} \right)(1-x+x^2)(2+x) - \left( \frac{1}{2+x} \right)(1-x+x^2)(2+x) \]

### Step 2: Simplify the right-hand side
The right-hand side simplifies as follows:

1. The first term becomes:
\[ x(2+x) = 2x + x^2 \]

2. The second term simplifies to:
\[ 1 - x + x^2 \]

So, we can rewrite the equation as:

\[ ax - 1 = (2x + x^2) - (1 - x + x^2) \]

### Step 3: Combine like terms
Now, simplify the right-hand side:

\[ ax - 1 = 2x + x^2 - 1 + x - x^2 \]

The \(x^2\) terms cancel out:

\[ ax - 1 = 3x - 1 \]

### Step 4: Compare coefficients
Now, we can compare coefficients on both sides of the equation:

\[ ax = 3x \]

This implies:

\[ a = 3 \]

### Conclusion
Thus, the value of \(a\) is

\[ \boxed{3} \]