VIP_19:11_催眠术ZERO kamma.2 「村越学

VIP_19:11_催眠术ZERO kamma.2 「村越学

由题意得：$\left\{\begin{array}{l}0\leqslant 11-2m\leqslant 5m\\0\leqslant 2m-2\leqslant 11-3m\end{array}\right.\Rightarrow \dfrac{11}{7}\leqslant m\leqslant \dfrac{13}{5}$；$\because m\in N$，$\therefore m=2$；

$\therefore a=\complement _{10}^{7}-A_{5}^{2}=100$；

$\because 77^{77}-14=(\left(19\times 4+1\right)^{77}-14=\complement _{77}^{0}\left(19\times 4\right)^{77}+\complement _{77}^{1}\left(19\times 4\right)^{76}+\ldots +\complement _{77}^{76}\left(19\times 4\right)+1-14$；

$\therefore b=6$；

$\left(\dfrac{a}{50}\sqrt {x}-\dfrac{1}{x}\right)^{b}=\left(2\sqrt {x}-\dfrac{1}{x}\right)^{6}$的展开式的通项公式为：$T_{r+1}=\complement _{6}^{r}\left(2\sqrt {x}\right)^{6-r}\cdot \left(-\dfrac{1}{x}\right)^{r}=\left(-1\right)^{r}\cdot 2^{6-r}\cdot \complement _{6}^{r}\cdot x^{\frac{6-3r}{2}}$；

令$\dfrac{6-3r}{2}=0\Rightarrow r=2$；

故其常数项为：$\left(-1\right)^{2}\cdot 2^{6-2}\cdot \complement _{6}^{2}=240$.