《2025年实验班提优训练七年级数学上册苏科版苏州专版》

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)$999×118\frac{4}{5}+999×(-\frac{1}{5})-999×18\frac{3}{5}.$

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

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

11. 整体思想 (2025·安徽淮北期末)阅读理解:
计算$(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})×(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5})-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5})×(\frac{1}{2}+\frac{1}{3}+\frac{1}{4})$时,若把$(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5})与(\frac{1}{2}+\frac{1}{3}+\frac{1}{4})$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$(\frac{1}{2}+\frac{1}{3}+\frac{1}{4})$为A,$(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5})$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$(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6})(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6});$
②$(1+\frac{1}{2}+\frac{1}{3}+... +\frac{1}{n})(\frac{1}{2}+\frac{1}{3}+... +\frac{1}{n+1})-(1+\frac{1}{2}+\frac{1}{3}+... +\frac{1}{n+1})(\frac{1}{2}+\frac{1}{3}+... +\frac{1}{n}).$

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