8. 请你参考黑板中老师的讲解，用运算律简便计算：
(1)$999×(-15)$；
(2)$999×118 \frac{4}{5} + 999×(-\frac{1}{5}) - 999×18 \frac{3}{5}$.
答案：8.(1)原式$=(1\ 000-1)×(-15)=15-15\ 000=-14\ 985.$
(2)原式$=999×[118\dfrac{4}{5}+(-\dfrac{1}{5})-18\dfrac{3}{5}]=999×100=99\ 900.$
9. 新考法 解题方法型阅读理解题 阅读下面材料：
$(1+\dfrac{1}{2})×(1-\dfrac{1}{3})=\dfrac{3}{2}×\dfrac{2}{3}=1,$
$(1+\dfrac{1}{2})×(1+\dfrac{1}{4})×(1-\dfrac{1}{3})×(1-\dfrac{1}{5})=\dfrac{3}{2}×\dfrac{5}{4}×\dfrac{2}{3}×\dfrac{4}{5}=\dfrac{3}{2}×\dfrac{2}{3}×\dfrac{5}{4}×\dfrac{4}{5}=1×1=1.$
根据以上信息，求出下式的结果.
$(1+\dfrac{1}{2})×(1+\dfrac{1}{4})×(1+\dfrac{1}{6})×···×(1+\dfrac{1}{20})×$
$(1-\dfrac{1}{3})×(1-\dfrac{1}{5})×(1-\dfrac{1}{7})×(1-\dfrac{1}{9})×···×$
$(1-\dfrac{1}{21}).$
答案：9.原式$=\dfrac{3}{2}×\dfrac{5}{4}×\dfrac{7}{6}×···×\dfrac{21}{20}×\dfrac{2}{3}×\dfrac{4}{5}×\dfrac{6}{7}×\dfrac{8}{9}×···×\dfrac{20}{21}=\dfrac{3}{2}×\dfrac{2}{3}×\dfrac{5}{4}×\dfrac{4}{5}×\dfrac{7}{6}×\dfrac{6}{7}×···×\dfrac{21}{20}×\dfrac{20}{21}=1×1×1×···×1=1.$
归纳总结 本题是一道阅读理解题,根据题目获取信息,利用获取的信息解决问题是解决这类题目的基本思路.
10. 整体思想(2025·安徽淮北期末)阅读理解:
计算$(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4})×(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5})-(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5})×(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4})$时,若把$(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5})$与$(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4})$分别各看作一个整体,再利用分配律进行运算,可以大大简化难度.过程如下:
解:设$(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4})$为$A$,$(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5})$为$B$,则原式$=B(1+A)-A(1+B)=B+AB - A - AB = B - A=\dfrac{1}{5}$.请用上面方法计算:
①$(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7})-(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7})(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})$;
②$(1+\dfrac{1}{2}+\dfrac{1}{3}+···+\dfrac{1}{n})(\dfrac{1}{2}+\dfrac{1}{3}+···+\dfrac{1}{n+1})-(1+\dfrac{1}{2}+\dfrac{1}{3}+···+\dfrac{1}{n+1})(\dfrac{1}{2}+\dfrac{1}{3}+···+\dfrac{1}{n})$.
答案：10. ①设$(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})$为$A$,$(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7})$为$B$,
原式$=(1+A)B-(1+B)A=B+AB-A-AB$
$\rightarrow$利用乘法分配律
$=B-A=\dfrac{1}{7}.$
②设$(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+···+\dfrac{1}{n})$为$A$,
$(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+···+\dfrac{1}{n+1})$为$B$,
原式$=(1+A)B-(1+B)A=B+AB-A-AB=B-A=\dfrac{1}{n+1}.$
