《2025年名校课堂七年级数学上册北师大版甘肃专版》

计算：
(1)$(-10)× (-7.32)× (-0.1)$.
(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： 解：
(1)原式=-(10×0.1)×7.32=-7.32.
(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$=[1\frac{2}{13}-(15-4+14)]÷(-5)=(\frac{15}{13}-25)×(-\frac{1}{5})=\frac{15}{13}×(-\frac{1}{5})-25×(-\frac{1}{5})=-\frac{3}{13}+5=\frac{10}{13}.(7)$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)$原式$=(0.7×1\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.$