《2025年名校课堂七年级数学上册北师大版安徽专版》

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)$9\frac{11}{12}×(-12)+35.5×4-5.5×4$.
(9)$0.7×1\frac{4}{9}+2\frac{3}{4}×(-17)+0.7×\frac{5}{9}+\frac{1}{4}×(-17)$.

答案： 针对训练
解：
(1)原式=-(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$=[1\frac{2}{13}-(15-4+14)]÷(-5)=(\frac{15}{13}-25)×(-\frac{1}{5})=$
$\frac{15}{13}×(-\frac{1}{5})-25×(-\frac{1}{5})=-\frac{3}{13}+5=\frac{10}{13}×（$原式最后计算结果应为$\frac{62}{13}，$推测原式计算过程$\frac{15}{13}×(-\frac{1}{5}) - 25×(-\frac{1}{5})$应计算错误，正确计算为$-\frac{3}{13}+5=\frac{-3 + 65}{13}=\frac{62}{13}，$但按原文识别输出）.
(7)因为$(\frac{2}{3}-\frac{1}{10}$
$+\frac{1}{6}-\frac{2}{5})÷(-\frac{1}{30})=(\frac{2}{3}-\frac{1}{10}+\frac{1}{6}-\frac{2}{5})×(-30)=\frac{2}{3}×$
$(-30)-\frac{1}{10}×(-30)+\frac{1}{6}×(-30)-\frac{2}{5}×(-30)=-20-$
(-3)+(-5)-(-12)=-10，所以$(\frac{2}{3}-\frac{1}{10}+\frac{1}{6}$
$-\frac{2}{5})÷(-\frac{1}{30})=-\frac{1}{10}×（$此处原式计算过程及结果可能存在错误，但按原文识别输出）.
(8)原式$=(10-\frac{1}{12})×(-12)+(35.5-5.5)×4=$
$10×(-12)-\frac{1}{12}×(-12)+30×4=-120+1+120=1.(9)$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.