If frac{ax - 1}{(1 - x + x^2)(2 + x)} = frac{ | vedclass

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(B) Given the equation: $\frac{ax - 1}{(1 - x + x^2)(2 + x)} = \frac{x}{1 - x + x^2} - \frac{1}{2 + x}$ Combine the terms on the right-hand side by ta

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$3$

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(B) Given the equation: $\frac{ax - 1}{(1 - x + x^2)(2 + x)} = \frac{x}{1 - x + x^2} - \frac{1}{2 + x}$ Combine the terms on the right-hand side by taking the common denominator: $\frac{x(2 + x) - 1(1 - x + x^2)}{(1 - x + x^2)(2 + x)} = \frac{2x + x^2 - 1 + x - x^2}{(1 - x + x^2)(2 + x)}$ Simplify the numerator: $\frac{3x - 1}{(1 - x + x^2)(2 + x)}$ Comparing this with the left-hand side $\frac{ax - 1}{(1 - x + x^2)(2 + x)}$,we get: $ax - 1 = 3x - 1$ Therefore,$a = 3$.

If $\frac{ax - 1}{(1 - x + x^2)(2 + x)} = \frac{x}{1 - x + x^2} - \frac{1}{2 + x}$,then $a = $

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