Segundo o enunciado, temos: \begin{matrix} b = \dfrac{1}{2}\cdot a \end{matrix}Assim:\begin{matrix} \left(a\cdot x - \dfrac{a}{2\cdot \sqrt{x}}\right)^{12} &\Rightarrow& a^{12}\cdot \left(x - \dfrac{1}{2\sqrt{x}}\right)^{12} & \text{(I)} \end{matrix}Conhecendo o $\text{Termo Geral do Binômio de Newton}$:\begin{matrix} (x+y)^n \Rightarrow T_{k+1} ={n \choose k} \cdot x^{n-k}\cdot y^k \end{matrix}Dessa forma, \begin{matrix} \left[x + \dfrac{(-1)}{2\sqrt{x}}\right]^{12} &\Rightarrow& T_{k+1} ={12 \choose k} \cdot x^{12-k}\cdot (-1)^k\cdot \left(\dfrac{x^{-1/2}}{2}\right)^k \end{matrix}Continuando,\begin{matrix} T_{k+1} ={12 \choose k} \cdot x^{12-3k/2}\cdot (-2^{-1})^k \end{matrix}Assim,\begin{matrix}12 -\dfrac{3k}{2} = 0 &\Rightarrow& k = 8 &\therefore& T_{9} ={12 \choose 8} {{\dfrac{1}{2^8}}} \end{matrix}Portanto, conforme $\text{(I)}$:\begin{matrix} a^{12}\cdot {12 \choose 8} \cdot {{\dfrac{1}{2^8}}} = 7920 &\Rightarrow& a^{12} =2^{12} &\Rightarrow & a = 2 \ , \ b =1 &\therefore& a+b =3 \end{matrix}\begin{matrix}Letra \ (B) \end{matrix}