If frac{x-2}{x^2(2x-3)} = frac{A}{x} + frac{B | vedclass

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(D) Given the partial fraction decomposition: $\frac{x-2}{x^2(2x-3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x-3}$.Multiplying both sides by $x^2(2x

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$B$

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(D) Given the partial fraction decomposition: $\frac{x-2}{x^2(2x-3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x-3}$.Multiplying both sides by $x^2(2x-3)$, we get: $x-2 = Ax(2x-3) + B(2x-3) + Cx^2$.Setting $x = 0$: $-2 = B(-3) \Rightarrow B = \frac{2}{3}$.Setting $x = \frac{3}{2}$: $\frac{3}{2} - 2 = C(\frac{3}{2})^2$ $\Rightarrow -\frac{1}{2} = C(\frac{9}{4})$ $\Rightarrow C = -\frac{2}{9}$.Comparing the coefficients of $x^2$: $0 = 2A + C$ $\Rightarrow 2A = -C = \frac{2}{9}$ $\Rightarrow A = \frac{1}{9}$.Now, calculate $2(A-C) = 2(\frac{1}{9} - (-\frac{2}{9})) = 2(\frac{1+2}{9}) = 2(\frac{3}{9}) = 2(\frac{1}{3}) = \frac{2}{3}$.Since $B = \frac{2}{3}$, we have $2(A-C) = B$.

If $\frac{x-2}{x^2(2x-3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x-3}$, then $2(A-C) = $

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