《2025年名校课堂七年级数学上册沪科版安徽专版》

计算：
(1)$(-20)×(-1\frac {1}{5})×(-0.3)×5$；
(2)$25×\frac {3}{4}-(-25)×\frac {1}{2}+25×(-\frac {1}{4})$；
(3)$(\frac {2}{3}-\frac {3}{4}+\frac {1}{6})÷(-\frac {1}{24})$；
(4)$(-36)×99\frac {71}{72}$；
(5)$[\frac {15}{13}-(\frac {5}{8}-\frac {1}{6}+\frac {7}{12})×24]÷(-5)$；
(6)$(-\frac {1}{30})÷(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})$；
(7)$0.7×1\frac {4}{9}+2\frac {3}{4}×(-17)+0.7×\frac {5}{9}+\frac {1}{4}×(-17)$。

答案： 针对训练
解:
(1)原式$=-20×\frac {6}{5}×\frac {3}{10}×5=-36$.
(2)原式$=25×\frac {3}{4}+25×\frac {1}{2}-25×\frac {1}{4}=25×(\frac {3}{4}+\frac {1}{2}-\frac {1}{4})=25×1=25$.
(3)原式$=(\frac {2}{3}-\frac {3}{4}+\frac {1}{6})×(-24)=-\frac {2}{3}×24+\frac {3}{4}×24-\frac {1}{6}×24=-16+18-4=-2$.
(4)解法一:原式$=(-36)×(100-\frac {1}{72})=-3600+\frac {1}{2}=-3599\frac {1}{2}$.解法二:原式$=-36×99\frac {71}{72}=-36×(100-\frac {1}{72})=-(36×100-36×\frac {1}{72})=-(3600-\frac {1}{2})=-3599\frac {1}{2}$.
(5)原式$=[\frac {15}{13}-(15-4+14)]÷(-5)=(\frac {15}{13}-25)×(-\frac {1}{5})=\frac {15}{13}×(-\frac {1}{5})-25×(-\frac {1}{5})=-\frac {3}{13}+5=\frac {62}{13}$.
(6)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$=-\frac {1}{10}$.
(7)原式$=(0.7×1\frac {4}{9}+0.7×\frac {5}{9})+[2\frac {3}{4}×(-17)+\frac {1}{4}×(-17)]=0.7×(1\frac {4}{9}+\frac {5}{9})+(-17)×(2\frac {3}{4}+\frac {1}{4})=0.7×2+(-17)×3=1.4-51=-49.6.$