frac{4x^2+5}{(x-2)^4} = frac{A}{(x-2)} + frac | vedclass

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(C) Let $x-2 = t$,so $x = t+2$. Substituting this into the expression: $4(t+2)^2 + 5 = 4(t^2 + 4t + 4) + 5 = 4t^2 + 16t + 21$. Now,$\frac{4t^2 + 16t +

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$\frac{5}{4}$

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(C) Let $x-2 = t$,so $x = t+2$. Substituting this into the expression: $4(t+2)^2 + 5 = 4(t^2 + 4t + 4) + 5 = 4t^2 + 16t + 21$. Now,$\frac{4t^2 + 16t + 21}{t^4} = \frac{4}{t^2} + \frac{16}{t^3} + \frac{21}{t^4}$. Comparing this with $\frac{A}{t} + \frac{B}{t^2} + \frac{C}{t^3} + \frac{D}{t^4}$,we get: $A = 0$,$B = 4$,$C = 16$,and $D = 21$. Now,calculate $\sqrt{\frac{A+B+D}{C}} = \sqrt{\frac{0+4+21}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

$\frac{4x^2+5}{(x-2)^4} = \frac{A}{(x-2)} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} + \frac{D}{(x-2)^4}$,then $\sqrt{\frac{A}{C} + \frac{B}{C} + \frac{D}{C}} = $

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