8[2026安徽安庆期中,中]计算:$(1+\frac{1}{2})×(1+\frac{1}{4})×(1+\frac{1}{6})×\dots×(1+\frac{1}{20})×(1-\frac{1}{3})×(1-\frac{1}{5})×(1-\frac{1}{7})×\dots×(1-\frac{1}{21}).$

原式$=\frac{3}{2}×\frac{5}{4}×\frac{7}{6}×\dots×\frac{21}{20}×\frac{2}{3}×\frac{4}{5}×\frac{6}{7}×\dots×\frac{20}{21}=(\frac{3}{2}×\frac{2}{3})×(\frac{5}{4}×\frac{4}{5})×(\frac{7}{6}×\frac{6}{7})×\dots×(\frac{21}{20}×\frac{20}{21})=1$.

9 用简便方法计算：
(1) [2025 山东济宁期中, 中] $(-2024 \frac{5}{6}) + 4046 \frac{2}{3} + (-2025 \frac{2}{3}) + 1 \frac{5}{6}$；
(2) [2025 河南郑州期中, 中] $(-199 \frac{37}{38}) × 76$。

(1) 原式$=[(-\ 2\ 024\ ) + (-\frac{5}{6})] + (\ 4\ 046 + \frac{2}{3}) + [(\ -\ 2\ 025\ ) + (-\frac{2}{3})] + (1 + \frac{5}{6}) = [(-2\ 024)+4\ 046+(-2\ 025)+1] + [(-\frac{5}{6})+\frac{2}{3}+(-\frac{2}{3})+\frac{5}{6}] = -2+0=-2$.
(2) 原式$=(-200+\frac{1}{38})×76=-200×76+\frac{1}{38}×76=-15\ 200+2=-15\ 198$.

10[难]阅读下面的解答过程.
计算:$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{9×10}$.
解:因为$\frac{1}{1×2}=1-\frac{1}{2}$,$\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3}$,$\frac{1}{3×4}=\frac{1}{3}-\frac{1}{4}$,$…$,$\frac{1}{9×10}=\frac{1}{9}-\frac{1}{10}$,
所以原式$=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+…+(\frac{1}{9}-\frac{1}{10})$
$=1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+…+(-\frac{1}{9}+\frac{1}{9})-\frac{1}{10}$
$=1-\frac{1}{10}$
$=\frac{9}{10}$.
根据以上解题方法计算:
(1)$\frac{1}{n(n+1)}=$($n$为正整数);
(2)$1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}$;
(3)$\frac{1}{2×4}+\frac{1}{4×6}+\frac{1}{6×8}+…+\frac{1}{2018×2020}$.

(2) 原式$=1-\frac{1}{1×2}-\frac{1}{2×3}-\frac{1}{3×4}-\frac{1}{4×5}-\frac{1}{5×6}-\frac{1}{6×7}=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\dots-\frac{1}{6}+\frac{1}{7}=\frac{1}{7}$.
(3) 原式$=\frac{1}{4}×(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{1\ 009}-\frac{1}{1\ 010})=\frac{1}{4}×(1-\frac{1}{1\ 010})=\frac{1}{4}×\frac{1\ 009}{1\ 010}=\frac{1\ 009}{4\ 040}$.

11[中]阅读下列材料，回答问题.
计算：$50÷(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{12})$.
解法1：原式$=50÷\dfrac{1}{3}-50÷\dfrac{1}{4}+50÷\dfrac{1}{12}=50×3-50×4+50×12$. 该解法对吗？答：.（填“对”或“不对”）
解法2：先计算原式的倒数，$(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{12})÷50=\dfrac{1}{3}×\dfrac{1}{50}-\dfrac{1}{4}×\dfrac{1}{50}+\dfrac{1}{12}×\dfrac{1}{50}=\dfrac{1}{300}$，故原式$=300$.
（1）请你用解法2的方法计算：$(-\dfrac{1}{30})÷(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})$；
（2）计算：$(1\dfrac{3}{4}-\dfrac{7}{8}-\dfrac{7}{12})÷(-\dfrac{7}{8})+(-\dfrac{7}{8})÷(1\dfrac{3}{4}-\dfrac{7}{8}-\dfrac{7}{12})$.

因为除法没有分配律,所以解法1不对.
(1) 先计算原式的倒数,$(\frac{2}{3}-\frac{1}{10}+\frac{1}{6}-\frac{2}{5})÷(-\frac{1}{30})=\frac{2}{3}×(-30)-\frac{1}{10}×(-30)+\frac{1}{6}×(-30)-\frac{2}{5}×(-30)=-20+3-5+12=-10$,故原式$=-\frac{1}{10}$.
(2) $(1\frac{3}{4}-\frac{7}{8}-\frac{7}{12})÷(-\frac{7}{8})=\frac{7}{4}×(-\frac{8}{7})-\frac{7}{8}×(-\frac{8}{7})-\frac{7}{12}×(-\frac{8}{7})=-2-(-1)-(-\frac{2}{3})=-2+1+\frac{2}{3}=-\frac{1}{3}$,所以 $(-\frac{7}{8})÷(1\frac{3}{4}-\frac{7}{8}-\frac{7}{12}) = -3$,所以原式 $= -\frac{1}{3} + (-3) = -\frac{10}{3}$.