If frac{2 x+7}{left(x^2+4right)left(x^2+9righ | vedclass

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(D) Given the partial fraction decomposition: $\frac{2 x+7}{\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}=\frac{A x+1}{x^2+4}+\frac{B x+m}{

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$\frac{109}{2}$

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(D) Given the partial fraction decomposition: $\frac{2 x+7}{\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}=\frac{A x+1}{x^2+4}+\frac{B x+m}{x^2+9}+\frac{C x+n}{x^2+16}$ Multiplying both sides by the denominator $\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)$,we get: $2 x+7 = (A x+1)(x^2+9)(x^2+16) + (B x+m)(x^2+4)(x^2+16) + (C x+n)(x^2+4)(x^2+9)$ Setting $x^2 = -4$: $2x + 7 = (Ax + 1)(5)(12) = 60Ax + 60$ Comparing coefficients of $x$: $60A = 2 \Rightarrow A = \frac{1}{30}$ Setting $x^2 = -9$: $2x + 7 = (Bx + m)(-5)(7) = -35Bx - 35m$ Comparing coefficients of $x$: $-35B = 2 \Rightarrow B = -\frac{2}{35}$ Setting $x^2 = -16$: $2x + 7 = (Cx + n)(-12)(-7) = 84Cx + 84n$ Comparing coefficients of $x$: $84C = 2 \Rightarrow C = \frac{2}{84} = \frac{1}{42}$ Now,calculating $\frac{1}{A} + \frac{1}{B} + \frac{1}{C}$: $\frac{1}{A} = 30$,$\frac{1}{B} = -\frac{35}{2}$,$\frac{1}{C} = 42$ $\frac{1}{A} + \frac{1}{B} + \frac{1}{C} = 30 - \frac{35}{2} + 42 = 72 - 17.5 = 54.5 = \frac{109}{2}$

If $\frac{2 x+7}{\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}=\frac{A x+1}{x^2+4}+\frac{B x+m}{x^2+9}+\frac{C x+n}{x^2+16}$,then $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=$

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