《2025年名校课堂七年级数学上册人教版2四川专版》

计算：
(1)$(-20)× (-1\frac {1}{5})× (-0.3)× 5$.
(2)$(\frac {5}{6}+\frac {7}{8}-\frac {23}{24})× (-48)$.
(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： 解:
(1)原式=-(20×0.3)×($\frac{6}{5}$×5)=-6×6=-36.
(2)原式=$\frac{5}{6}$×(-48)+$\frac{7}{8}$×(-48)-$\frac{23}{24}$×(-48)=-40-42+46=-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)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(-$\frac{1}{30}$)÷($\frac{2}{3}$-$\frac{1}{10}$+$\frac{1}{6}$-$\frac{2}{5}$)=-$\frac{1}{10}$.
(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.