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

【解】$(1+\dfrac{1}{2}) × (1+\dfrac{1}{4}) × (1+\dfrac{1}{6}) × \dots × (1+\dfrac{1}{20}) × (1-\dfrac{1}{3}) × (1-\dfrac{1}{5}) × (1-\dfrac{1}{7}) × \dots × (1-\dfrac{1}{21}) =\dfrac{3}{2} × \dfrac{5}{4} × \dfrac{7}{6} × \dots × \dfrac{21}{20} × \dfrac{2}{3} × \dfrac{4}{5} × \dfrac{6}{7} × \dots × \dfrac{20}{21} = ( \dfrac{3}{2} × \dfrac{2}{3} ) × ( \dfrac{5}{4} × \dfrac{4}{5} ) × ( \dfrac{7}{6} × \dfrac{6}{7} ) × \dots × ( \dfrac{21}{20} × \dfrac{20}{21} ) =1.$

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] $(-199 \frac{37}{38}) × 76$。

【解】(1) $(-2\ 024\ \dfrac{5}{6}) + 4\ 046\ \dfrac{2}{3} + (-2\ 025\ \dfrac{2}{3}) +1\ \dfrac{5}{6} = [ (-2\ 024) + (-\dfrac{5}{6}) ] + ( 4\ 046 + \dfrac{2}{3} ) + [ (-2\ 025) + (-\dfrac{2}{3}) ] + ( 1 + \dfrac{5}{6} ) = [ (-2\ 024) + 4\ 046 + (-2\ 025) + 1 ] + [ (-\dfrac{5}{6}) + \dfrac{2}{3} + (-\dfrac{2}{3}) + \dfrac{5}{6} ] =-2+0=-2.$
(2) $(-199\ \dfrac{37}{38}) × 76 = (-200 + \dfrac{1}{38}) × 76 =-200×76+\dfrac{1}{38}×76=-15\ 200+2=-15\ 198.$

10[难]阅读下面的解答过程.
计算: $\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\dots+\frac{1}{9×10}$.
解:因为$\frac{1}{1×2}=1-\frac{1}{2}$,$\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3}$,$\frac{1}{3×4}=\frac{1}{3}-\frac{1}{4}$,$\dots$,$\frac{1}{9×10}=\frac{1}{9}-\frac{1}{10}$,
所以原式$=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\dots+(\frac{1}{9}-\frac{1}{10})$
$=1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+\dots+(-\frac{1}{9}+\frac{1}{9})-\frac{1}{10}$
$=1-\frac{1}{10}$
$=\frac{9}{10}$.
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)$\frac{1}{2×4}+\frac{1}{4×6}+\frac{1}{6×8}+\dots+\frac{1}{2018×2020}$.

(1)$\frac{1}{n}-\frac{1}{n+1}$
【解】(2)$1-\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}=1-\dfrac{1}{1×2}-\dfrac{1}{2×3}-\dfrac{1}{3×4}-\dfrac{1}{4×5}-\dfrac{1}{5×6}-\dfrac{1}{6×7}=1-1+\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dots-\dfrac{1}{6}+\dfrac{1}{7}=\dfrac{1}{7}.$
(3)原式$=\dfrac{1}{4}×( 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{1\ 009}-\dfrac{1}{1\ 010} ) =\dfrac{1}{4} × ( 1-\dfrac{1}{1\ 010} ) =\dfrac{1}{4} × \dfrac{1\ 009}{1\ 010} =\dfrac{1\ 009}{4\ 040}.$

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: $( 1\dfrac{3}{4} - \dfrac{7}{8} - \dfrac{7}{12} ) ÷ ( -\dfrac{7}{8} ) + ( -\dfrac{7}{8} ) ÷ ( 1\dfrac{3}{4} - \dfrac{7}{8} - \dfrac{7}{12} )$.

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