For any quadratic polynomial f(x), it is true that $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{a}\right)+{\mathrm{f}}^{\text{'}}\left(\mathrm{a}\right)\left(\mathrm{x}-\mathrm{a}\right)+\frac{{\mathrm{f}}^{\text{'}\text{'}}\left(\mathrm{a}\right)}{2!}{\left(\mathrm{x}-\mathrm{a}\right)}^2$where a is any real number . If

$\frac{3{\mathrm{x}}^2+4\mathrm{x}+7}{{\left(\mathrm{x}-2\right)}^3}=\frac{\mathrm{A}}{{\left(\mathrm{x}-2\right)}^3}+\frac{\mathrm{B}}{{\left(\mathrm{x}-2\right)}^2}+\frac{\mathrm{C}}{\left(\mathrm{x}-2\right)} \mathrm{and} \mathrm{g}\left(\mathrm{x}\right)=3{\mathrm{x}}^2+4\mathrm{x}+7 \mathrm{then} \mathrm{A}+\mathrm{B}+\mathrm{C}=$

$\begin{array}{rcl} & & \frac{3{\mathrm{x}}^2+4\mathrm{x}+7}{{(\mathrm{x}-2)}^3}\\ \mathrm{x}-2& =& \mathrm{y} \Rightarrow \mathrm{x}=\mathrm{y}+2\\ & & \frac{3{(\mathrm{y}+2)}^2+4{(\mathrm{y}+2)}^{\mathrm{y}}}{{\mathrm{y}}^3}\\ & =& \frac{3{\mathrm{y}}^2+16\mathrm{y}+27}{{\mathrm{y}}^3}\\ & =& \frac{3}{\mathrm{y}}+\frac{16}{{\mathrm{y}}^2}+\frac{27}{{\mathrm{y}}^3}=\frac{27}{{(\mathrm{x}-2)}^3}+\frac{16}{{(\mathrm{x}-2)}^2}+\frac{3}{\mathrm{x}-2}\\ \mathrm{A}& =& 27, \mathrm{B}=16, \mathrm{C}=3\\ & & \begin{array}{rcl}\mathrm{A}+\mathrm{B}+\mathrm{C}& =& 46\end{array}\end{array}$

$$\begin{gathered} \mathrm{g}\left(\mathrm{x}\right)=3{\mathrm{x}}^2+4\mathrm{x}+7 \\ {\mathrm{g}}^1\left(\mathrm{x}\right)=6\mathrm{x}+4 \\ {\mathrm{g}}^{11}\left(\mathrm{x}\right)=6 \\ \mathrm{g}\left(2\right)=27, {\mathrm{g}}^1\left(2\right)=16, \frac{{\mathrm{g}}^{11}\left(2\right)}{2}=3 \\ \mathrm{g}\left(2\right)+\frac{{\mathrm{g}}^1\left(2\right)}{2}+\frac{{\mathrm{g}}^{11}\left(2\right)}{2}=46 \end{gathered}$$