《2025年名校课堂七年级数学上册人教版福建专版》

计算：
(1)$(-\frac {9}{7})÷[(-\frac {3}{14})×(-\frac {3}{2})]$.
(2)$(-20)×(-1\frac {1}{5})×(-0.3)×5$.
(3)$25×\frac {3}{4}-(-25)×\frac {1}{2}+25×(-\frac {1}{4})$.
(4)$(\frac {2}{3}-\frac {3}{4}+\frac {1}{6})÷(-\frac {1}{24})$.
(5)$(-36)×99\frac {71}{72}$.
(6)$[1\frac {2}{13}-(\frac {5}{8}-\frac {1}{6}+\frac {7}{12})×24]÷(-5)$.
(7)$(-\frac {1}{30})÷(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})$.
(8)$-|-\frac {1}{30}|-|-\frac {2}{3}+\frac {3}{5}|-|-\frac {1}{3}×\frac {1}{5}|-|-2|$.

答案：
(1)原式=$(-\frac {9}{7})÷(-\frac {3}{14})÷(-\frac {3}{2})=(-\frac {9}{7})×(-\frac {14}{3})×(-\frac {2}{3})=-4$.
(2)原式$=-(20×0.3)×(\frac {6}{5}×5)=-6×6=-36$.
(3)原式$=25×\frac {3}{4}+25×\frac {1}{2}-25×\frac {1}{4}=25×(\frac {3}{4}+\frac {1}{2}-\frac {1}{4})=25×1=25$.
(4)原式$=(\frac {2}{3}-\frac {3}{4}+\frac {1}{6})×(-24)=-\frac {2}{3}×24+\frac {3}{4}×24-\frac {1}{6}×24=-16+18-4=-2$.
(5)原式$=(-36)×(100-\frac {1}{72})=-3600+\frac {1}{2}=-3599\frac {1}{2}$.
(6)原式$=[1\frac {2}{13}-(15-4+14)]÷(-5)=(\frac {15}{13}-25)×(-\frac {1}{5})=\frac {15}{13}×(-\frac {1}{5})-25×(-\frac {1}{5})=-\frac {3}{13}+5=4\frac {10}{13}$.
(7)原式的倒数为$(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})÷(-\frac {1}{30})=(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})×(-30)=\frac {2}{3}×(-30)-\frac {1}{10}×(-30)+\frac {1}{6}×(-30)-\frac {2}{5}×(-30)=-20-(-3)+(-5)-(-12)=-10$.故原式$=-\frac {1}{10}$.
(8)原式$=-\frac {1}{30}-\frac {1}{15}-\frac {1}{15}-2=-\frac {1}{6}-2=-2\frac {1}{6}$.