《2025年名校课堂七年级数学上册湘教版广西专版》

计算：
(1)$(-13)×(-134)×\frac {1}{13}×(-\frac {1}{67})$；
(2)$54×(-95)+38×(-95)-8×95$；
(3)$(-\frac {1}{3}-\frac {1}{4}+\frac {1}{5}-\frac {7}{15})×(-60)$；
(4)$(-36)×99\frac {71}{72}$；
(5)$(\frac {1}{3}+\frac {1}{6}-\frac {2}{5})÷(-\frac {1}{30})$；
(6)$\frac {1}{12}÷(-\frac {1}{6}-\frac {2}{3}+\frac {1}{4})$；
(7)$(\frac {7}{9}-\frac {5}{6}+\frac {3}{18})×18+3.95×6-1.45×6$；
(8)$0.3×2\frac {5}{11}+3\frac {4}{5}×(-7)+0.3×\frac {6}{11}+\frac {1}{5}×(-7)$。

答案：
(1)解:原式=-13×134×$\frac{1}{13}$×$\frac{1}{67}$=-(13×$\frac{1}{13}$)×(134×$\frac{1}{67}$)=-1×2=-2.
(2)解:原式=(-95)×(54+38+8)=-95×100=-9500.
(3)解:原式=(-$\frac{1}{4}$)×(-60)-$\frac{1}{3}$×(-60)+$\frac{1}{5}$×(-60)-$\frac{7}{15}$×(-60)=20+15-12+28=51.
(4)解:原式=(-36)×(100-$\frac{1}{72}$)=-3600+$\frac{1}{2}$=-3599$\frac{1}{2}$.
(5)解:原式=$\frac{1}{3}$×(-30)+$\frac{1}{6}$×(-30)-$\frac{2}{5}$×(-30)=-10+(-5)-(-12)=-10-5+12=-3.
(6)解:因为(-$\frac{1}{6}$-$\frac{2}{3}$+$\frac{1}{4}$)÷$\frac{1}{12}$=(-$\frac{1}{6}$-$\frac{2}{3}$+$\frac{1}{4}$)×12=-$\frac{1}{6}$×12-$\frac{2}{3}$×12+$\frac{1}{4}$×12=-2-8+3=-7,所以$\frac{1}{12}$÷(-$\frac{1}{6}$-$\frac{2}{3}$+$\frac{1}{4}$)=-$\frac{1}{7}$.
(7)解:原式=$\frac{7}{9}$×18-$\frac{5}{6}$×18+$\frac{3}{18}$×18+(3.95-1.45)×6=14-15+3+2.5×6=2+15=17.
(8)解:原式=(0.3×2$\frac{5}{11}$+0.3×$\frac{6}{11}$)+[3$\frac{4}{5}$×(-7)+$\frac{1}{5}$×(-7)]=0.3×(2$\frac{5}{11}$+$\frac{6}{11}$)+(-7)×(3$\frac{4}{5}$+$\frac{1}{5}$)=0.3×3+(-7)×4=0.9-28=-27.1.