《2025年名校课堂七年级数学上册人教版山西专版》

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)$0.7×1\frac {4}{9}+2\frac {3}{4}×(-17)+0.7×\frac {5}{9}+\frac {1}{4}×(-17)$.

答案：
(1)原式$=( - \frac{9}{7})÷\frac{9}{28}=( - \frac{9}{7})×\frac{28}{9}= - 4.(2)$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$=36×[( - 100) + \frac{1}{72}]=36×( - 100) + 36×\frac{1}{72}= - 3600 + \frac{1}{2}= - 3599\frac{1}{2}.(6)$原式$=[1\frac{2}{13} - (\frac{5}{8}×24 - \frac{1}{6}×24 + \frac{7}{12}×24)]÷( - 5)=[1\frac{2}{13} - (15 - 4 + 14)]÷( - 5)=(\frac{15}{13} - 25)×( - \frac{1}{5})=\frac{15}{13}×( - \frac{1}{5}) - 25×( - \frac{1}{5})= - \frac{3}{13} + 5=4\frac{10}{13}.(7)$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$=(0.7×\frac{4}{9} + 0.7×\frac{5}{9}) + [2\frac{3}{4}×( - 17) + \frac{1}{4}×( - 17)]=0.7×(\frac{4}{9} + \frac{5}{9}) + ( - 17)×(2\frac{3}{4} + \frac{1}{4})=0.7×2 + ( - 17)×3=1.4 - 51= - 49.6.$