If we resolve the rational fraction frac{1}{( | vedclass

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(D) Given the partial fraction decomposition: $\frac{1}{(1-2x)^2(1-3x)} = \frac{A}{1-3x} + \frac{B}{1-2x} + \frac{C}{(1-2x)^2}$.Multiplying both sides

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$-6$

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(D) Given the partial fraction decomposition: $\frac{1}{(1-2x)^2(1-3x)} = \frac{A}{1-3x} + \frac{B}{1-2x} + \frac{C}{(1-2x)^2}$.Multiplying both sides by $(1-2x)^2(1-3x)$,we get: $1 = A(1-2x)^2 + B(1-2x)(1-3x) + C(1-3x)$.Setting $x = \frac{1}{3}$: $1 = A(1 - \frac{2}{3})^2 = A(\frac{1}{3})^2 = \frac{A}{9} \implies A = 9$.Setting $x = \frac{1}{2}$: $1 = C(1 - \frac{3}{2}) = C(-\frac{1}{2}) \implies C = -2$.Setting $x = 0$: $1 = A(1)^2 + B(1)(1) + C(1) = A + B + C$.Substituting $A = 9$ and $C = -2$: $1 = 9 + B - 2 \implies 1 = 7 + B \implies B = -6$.Thus,the set is $\{9, -6, -2\}$.The minimum value is $\min \{9, -6, -2\} = -6$.

If we resolve the rational fraction $\frac{1}{(1-2x)^2(1-3x)}$ into partial fractions of the form $\frac{A}{1-3x} + \frac{B}{1-2x} + \frac{C}{(1-2x)^2}$,then what is $\min \{A, B, C\} = $

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