9. 计算：
(1)$25×\dfrac{1}{5}+25×\dfrac{1}{10}-25×\dfrac{1}{2}$；
(2)$(-48)×0.125+48×\dfrac{1}{8}+(-48)×\dfrac{5}{4}$；
(3)$(-5)×(-3\dfrac{6}{7})+(-7)×(-3\dfrac{6}{7})+12×(-3\dfrac{6}{7})$；
(4)$-6×\dfrac{3}{7}+4×\dfrac{3}{7}-5×\dfrac{3}{7}$；
(5)$99\dfrac{17}{18}×(-9)$；
(6)$49\dfrac{24}{25}×(-5)$。

9.解:(1)$原式=25×(\dfrac{1}{5}+\dfrac{1}{10}-\dfrac{1}{2})=25×(-\dfrac{1}{5})=-5$.
(2)$原式=(-48)×(0.125-\dfrac{1}{8}+\dfrac{5}{4})=(-48)×\dfrac{5}{4}=-60$.
(3)$原式=(-5-7+12)×(-3\dfrac{6}{7})=0×(-3\dfrac{6}{7})=0$.
(4)$原式=\dfrac{3}{7}×(-6+4-5)=\dfrac{3}{7}×(-7)=-3$.
(5)$原式=(100-\dfrac{1}{18})×(-9)=-900+\dfrac{1}{2}=-899\dfrac{1}{2}$.
(6)$原式=(50-\dfrac{1}{25})×(-5)=-250+\dfrac{1}{5}=-249\dfrac{4}{5}$.

10. 若 $a,b$ 互为相反数，$c$ 的倒数是 $4$，$d$ 的绝对值是最小的正整数.
求: (1)$3a+3b-4c$ 的值；
(2)$8c-d+cd$ 的值.

10.解:由题意,得$a+b=0,c=\dfrac{1}{4},|d|=1$,所以$d=\pm1$.
(1)$3a+3b-4c=3(a+b)-4c=3×0-4×\dfrac{1}{4}=0-1=-1$.
(2)当$c=\dfrac{1}{4},d=1$时,$8c-d+cd=8×\dfrac{1}{4}-1+\dfrac{1}{4}×1=2-1+\dfrac{1}{4}=\dfrac{5}{4}$;
当$c=\dfrac{1}{4},d=-1$时,$8c-d+cd=8×\dfrac{1}{4}-(-1)+\dfrac{1}{4}×(-1)=2+1-\dfrac{1}{4}=\dfrac{11}{4}$.
综上,$8c-d+cd$的值为$\dfrac{5}{4}$或$\dfrac{11}{4}$.

11. 我们知道: $\dfrac{1}{2} × \dfrac{2}{3} = \dfrac{1}{3}$, $\dfrac{1}{2} × \dfrac{2}{3} × \dfrac{3}{4} = \dfrac{1}{4}$, $\dfrac{1}{2} × \dfrac{2}{3} × \dfrac{3}{4} × \dfrac{4}{5} = \dfrac{1}{5}$, $\dots$, $\dfrac{1}{2} × \dfrac{2}{3} × \dfrac{3}{4} × \dots × \dfrac{n}{n+1} = \dfrac{1}{n+1}$.
试根据以上规律,解答下列问题:
(1)计算: $(\dfrac{1}{2} - 1) × (\dfrac{1}{3} - 1) × (\dfrac{1}{4} - 1) × \dots × (\dfrac{1}{100} - 1)$;
(2)将 2026 减去它的$\dfrac{1}{2}$,再减去余下的$\dfrac{1}{3}$,再减去余下的$\dfrac{1}{4}$,再减去余下的$\dfrac{1}{5}$,$\dots\dots$,依此类推,直至减去余下的$\dfrac{1}{2026}$,最后的结果是多少?

11.解:(1)$原式=(-\dfrac{1}{2})×(-\dfrac{2}{3})×(-\dfrac{3}{4})×\dots×(-\dfrac{99}{100})=-\dfrac{1}{100}$.
(2)$2026×(1-\dfrac{1}{2})×(1-\dfrac{1}{3})×\dots×(1-\dfrac{1}{2026})=2026×\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{3}{4}×\dots×\dfrac{2025}{2026}=2026×\dfrac{1}{2026}=1$.