If frac{x-4}{x^2-5x+6} can be expanded in the | vedclass

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Question

(A) First,express the fraction using partial fractions: $\frac{x-4}{(x-2)(x-3)} = \frac{2}{x-2} - \frac{1}{x-3}$.Rewrite the terms to facilitate binom

Vedclass · Accepted Answer

$\frac{-73}{648}$

Vedclass · Answer

(A) First,express the fraction using partial fractions: $\frac{x-4}{(x-2)(x-3)} = \frac{2}{x-2} - \frac{1}{x-3}$.Rewrite the terms to facilitate binomial expansion: $2(x-2)^{-1} - (x-3)^{-1} = 2(-2)^{-1}(1-\frac{x}{2})^{-1} - (-3)^{-1}(1-\frac{x}{3})^{-1}$.This simplifies to: $-(1-\frac{x}{2})^{-1} + \frac{1}{3}(1-\frac{x}{3})^{-1}$.Using the expansion $(1-y)^{-1} = 1 + y + y^2 + y^3 + \dots$,we get:$-(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \dots) + \frac{1}{3}(1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} + \dots)$.The coefficient of $x^3$ is $-\frac{1}{8} + \frac{1}{3} 	imes \frac{1}{27} = -\frac{1}{8} + \frac{1}{81}$.Calculating the sum: $\frac{-81 + 8}{648} = -\frac{73}{648}$.

If $\frac{x-4}{x^2-5x+6}$ can be expanded in the ascending powers of $x$,then the coefficient of $x^3$ is

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