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

Given that

$\frac{2x+7}{(x^2+4)(x^2+9)(x^2+16)}=\frac{Ax+1}{x^2+4}+\frac{Bx+m}{x^2+9}+\frac{Cx+n}{x^2+16}$

$$\begin{gathered} \frac{2x+7}{(x^2+4)\left(x^2+9\right)(x^2+16)} = \\ \frac{(Ax+1)(x^2+9)(x^2+16)+(Bx+m)(x^2+4)(x^2+16) \\ +(Cx+n)(x^2+4)(x^2+9)}{(x^2+4)\left(x^2+9\right)(x^2+16)} \end{gathered}$$

Now

$$\begin{gathered} 2x+7=(Ax+1)(x^2+9)(x^2+16)+(Bx+m)(x^2+4) \\ (x^2+16)+(Cx+n)(x^2+4)(x^2+9) \end{gathered}$$

Now put $x^2=-4$

$2x+7=(Ax+1)(-4+9)(-4+16)+B\left(0\right)+C\left(0\right)$

$2x+7=(Ax+1)60$

Comparing both sides $x$ coefficient

$60A=2\Rightarrow A=\frac{1}{30}$

Now Put $x^2=-9$

$2x+7=A\left(0\right)+(Bx+m)(-5)\left(7\right)+C\left(0\right)$

$2x+7=(Bx+m)(-35)$

Comparing both sides $x$ coefficient

$2=-35B$

$\Rightarrow B=\frac{-2}{35}$

Now put $x^2=-16$

$2x+7=A\left(0\right)+B\left(0\right)+(Cx+n)(-12)(-7)$

$2x+7=(Cx+n)\left(84\right)$

$\Rightarrow C=\frac{2}{84}$

Now $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=30-\frac{35}{2}+\frac{84}{2}$

$=\frac{109}{2}$