8 [较难] 计算：$(-\dfrac{1}{2}×\dfrac{3}{2})×(-\dfrac{2}{3}×\dfrac{4}{3})×(-\dfrac{3}{4}×\dfrac{5}{4})×···×(-\dfrac{2021}{2022}×\dfrac{2023}{2022}).$

原式$=-\dfrac{1}{2}×(\dfrac{3}{2}×\dfrac{2}{3})×(\dfrac{4}{3}×\dfrac{3}{4})×(\dfrac{5}{4}×\dfrac{4}{5})×\dots×(\dfrac{2022}{2021}×\dfrac{2021}{2022})×\dfrac{2023}{2022} = -\dfrac{1}{2}×1×1×1×\dots×1×\dfrac{2023}{2022} = -\dfrac{2023}{4044}$.

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) $(-2024\dfrac{5}{6})+4046\dfrac{2}{3}+(-2025\dfrac{2}{3})+1\dfrac{5}{6} = [(-2024)+(-\dfrac{5}{6})]+(4046+\dfrac{2}{3})+[(-2025)+(-\dfrac{2}{3})]+(1+\dfrac{5}{6}) = [(-2024)+4046+(-2025)+1]+[(-\dfrac{5}{6})+\dfrac{2}{3}+(-\dfrac{2}{3})+\dfrac{5}{6}] = -2+0 = -2$.
(2) $(-199\dfrac{37}{38})×76 = (-200+\dfrac{1}{38})×76 = -200×76+\dfrac{1}{38}×76 = -15200+2 = -15198$.

10[2026上海杨浦区期中,较难]$\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}-\frac{21}{110}$
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原式$=\dfrac{5}{6}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{10}-\dfrac{1}{10}-\dfrac{1}{11} = \dfrac{5}{6}+\dfrac{1}{3}-\dfrac{1}{11} = \dfrac{77}{66}-\dfrac{6}{66} = \dfrac{71}{66}$.

11[2026安徽亳州质检,中]老师为了强化同学们的运算思维,提高数学运算能力,布置了一道有意思的计算题,鼓励大家用不同的方法计算$\frac{1}{24}÷(\frac{1}{3}-\frac{1}{4}+\frac{1}{12})$.以下是三名同学的计算过程.
甲:原式=$\frac{1}{24}÷\frac{1}{3}-\frac{1}{24}÷\frac{1}{4}+\frac{1}{24}÷\frac{1}{12}=\frac{1}{24}×3-\frac{1}{24}×4+\frac{1}{24}×12=\frac{11}{24}$.
乙:原式=$\frac{1}{24}÷(\frac{4}{12}-\frac{3}{12}+\frac{1}{12})=\frac{1}{24}÷\frac{1}{6}=\frac{1}{24}×6=\frac{1}{4}$.
丙:原式的倒数=$(\frac{1}{3}-\frac{1}{4}+\frac{1}{12})÷\frac{1}{24}=(\frac{1}{3}-\frac{1}{4}+\frac{1}{12})×24=\frac{1}{3}×24-\frac{1}{4}×24+\frac{1}{12}×24=4$,所以原式=$\frac{1}{4}$.
(1)比较他们的做法,其中的做法是错误的.
(2)选择合适的方法计算:$(-\frac{1}{210})÷(\frac{3}{7}+\frac{2}{30}-\frac{3}{10}-\frac{5}{21})$.
(3)直接写出$(\frac{3}{7}+\frac{2}{30}-\frac{3}{10}-\frac{5}{21})÷(-\frac{1}{210})×[(-\frac{1}{210})÷(\frac{3}{7}+\frac{2}{30}-\frac{3}{10}-\frac{5}{21})]$的结果.

(1) 由甲、乙、丙的做法可知,甲的做法是错误的.
(2) 因为$(\dfrac{3}{7}+\dfrac{2}{30}-\dfrac{3}{10}-\dfrac{5}{21})÷(-\dfrac{1}{210}) = (\dfrac{3}{7}+\dfrac{2}{30}-\dfrac{3}{10}-\dfrac{5}{21})×(-210) = \dfrac{3}{7}×(-210)+\dfrac{2}{30}×(-210)-\dfrac{3}{10}×(-210)-\dfrac{5}{21}×(-210) = -90-14+63+50 = 9$,所以$(-\dfrac{1}{210})÷(\dfrac{3}{7}+\dfrac{2}{30}-\dfrac{3}{10}-\dfrac{5}{21}) = \dfrac{1}{9}$.
(3) 由(2)可得原式$=1$.