《2025年通城学典初中数学运算能手七年级上册人教版》

12. (27分)计算：
(1)$-16-3^{2}÷(-2+\frac {1}{2})$ (2)$(-81)÷(+3\frac {1}{4})×(-\frac {4}{9})÷(-1\frac {1}{13})$
(3)$-270×\frac {1}{4}+0.25×21.5+(-8\frac {1}{2})×(-25\% )$

答案： （1）-10 （2）$-\frac{72}{7}$ （3）-60

13. (14分)观察下列各式：
$\frac {1}{2}×\frac {2}{3}= \frac {1}{3}$,$\frac {1}{2}×\frac {2}{3}×\frac {3}{4}= \frac {1}{4}$,$\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×\frac {4}{5}= \frac {1}{5}$,…
(1)猜想：$\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×... ×\frac {49}{50}= $$\frac{1}{50}$；
(2)根据上面的规律,计算：$(\frac {1}{100}-1)×(\frac {1}{99}-1)×(\frac {1}{98}-1)×... ×(\frac {1}{2}-1)$.
原式$=-\frac{99}{100}×\left(-\frac{98}{99}\right)×\left(-\frac{97}{98}\right)×\cdots×\left(-\frac{1}{2}\right)=-\frac{1}{100}$

答案： （1）$\frac{1}{50}$ （2）原式$=-\frac{99}{100}×\left(-\frac{98}{99}\right)×\left(-\frac{97}{98}\right)×\cdots×\left(-\frac{1}{2}\right)=-\frac{1}{100}$

14. (15分)我们知道$a÷b= \frac {a}{b}$,$b÷a= \frac {b}{a}$,显然$a÷b与b÷a$的结果互为倒数.小明利用这一思想方法计算$(-\frac {1}{30})÷(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})$的过程如下：因为$(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})÷(-\frac {1}{30})= (\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})×(-30)= -20+3-5+12= -10$,所以原式$=-\frac {1}{10}$.请你仿照这种方法计算：$(-\frac {1}{42})÷(\frac {1}{6}-\frac {3}{14}+\frac {2}{3}-\frac {2}{7})$.

答案： 因为$\left(\frac{1}{6}-\frac{3}{14}+\frac{2}{3}-\frac{2}{7}\right)÷\left(-\frac{1}{42}\right)=\left(\frac{1}{6}-\frac{3}{14}+\frac{2}{3}-\frac{2}{7}\right)×(-42)=\frac{1}{6}×(-42)-\frac{3}{14}×(-42)+\frac{2}{3}×(-42)-\frac{2}{7}×(-42)=-7+9-28+12=-14$,所以$\left(-\frac{1}{42}\right)÷\left(\frac{1}{6}-\frac{3}{14}+\frac{2}{3}-\frac{2}{7}\right)=-\frac{1}{14}$