《2025年实验班提优训练七年级数学上册苏科版》

9. 请你参考黑板中老师的讲解,用运算律简便计算：
利用运算律有时能进行简便计算.
例1:$98×12= (100-2)×12= 1200-24= 1176$；
例2:$-16×233+17×233= (-16+17)×233= 233$.
(第9题)
(1)$999×(-15)$；
(2)$999×118\frac{4}{5}+999×\left(-\frac{1}{5}\right)-999×18\frac{3}{5}$.

答案：
(1)原式=$(1000-1)×(-15)=15-15000=-14985$.
(2)原式=$999×\left[118\frac{4}{5}+(-\frac{1}{5})-18\frac{3}{5}\right]=999×100=99900$.

10. 我们知道：$\frac{1}{2}×\frac{2}{3}= \frac{1}{3}$,$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}= \frac{1}{4}$,$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}×\frac{4}{5}= \frac{1}{5}$,…$$,$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}×…×\frac{n}{n+1}= \frac{1}{n+1}$. 试根据上面的规律,解答下列各题：
(1)计算：$\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)…\left(\frac{1}{100}-1\right)$.
(2)将2025减去它的$\frac{1}{2}$,再减去余下的$\frac{1}{3}$,再减去余下的$\frac{1}{4}$,再减去余下的$\frac{1}{5}$,…$$,依次类推,直到最后减去余下的$\frac{1}{2025}$,最后的结果是多少？

答案：
(1)原式=$(-\frac{1}{2})×(-\frac{2}{3})×(-\frac{3}{4})×\cdots×(-\frac{99}{100})=-\frac{1}{2}×\frac{2}{3}×\frac{3}{4}×\cdots×\frac{99}{100}=-\frac{1}{100}$.
(2)根据题意,得$2025×(1-\frac{1}{2})×(1-\frac{1}{3})×\cdots×(1-\frac{1}{2025})=2025×\frac{1}{2}×\frac{2}{3}×\cdots×\frac{2024}{2025}=1$.

11. 整体思想 (2025·安徽淮北期末)阅读理解：
计算$\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)×\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)×\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)$时,若把$\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)与\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)$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$\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)为A$,$\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)为B$,则原式$=B(1+A)-A(1+B)= B+AB-A-AB= B-A= \frac{1}{5}$. 请用上面方法计算：
①$\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)$；
②$\left(1+\frac{1}{2}+\frac{1}{3}+…+\frac{1}{n}\right)\left(\frac{1}{2}+\frac{1}{3}+…+\frac{1}{n+1}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+…+\frac{1}{n+1}\right)\left(\frac{1}{2}+\frac{1}{3}+…+\frac{1}{n}\right)$.

答案： ①设$(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6})$为A,$(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})$为B,原式=$(1+A)B-(1+B)A=B+AB-A-AB$$\to$利用乘法分配律$=B-A=\frac{1}{7}$.②设$(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots+\frac{1}{n})$为A,$(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\cdots+\frac{1}{n+1})$为B,原式=$(1+A)B-(1+B)A=B+AB-A-AB=B-A=\frac{1}{n+1}$.