《2025年课堂点睛七年级数学上册北师大版》

1. 计算:$4\frac {3}{4}-(+3.85)-3\frac {1}{4}+(-3.15).$

答案： 解:原式$=\left(4\frac{3}{4}-3\frac{1}{4}\right)+(-3.85-3.15)=1\frac{1}{2}-7=-5.5$.

2. 计算:$(-1.25)×2.5÷5÷\frac {1}{4}×8.$

答案： 解:原式$=-\frac{5}{4}×\frac{5}{2}×\frac{1}{5}×4×8=\left(-\frac{5}{4}×4\right)×\left(\frac{5}{2}×\frac{1}{5}×8\right)=-5×4=-20$.

3. 计算:$(\frac {3}{4}-\frac {5}{6}+\frac {7}{12})×(-4)×(-3)^{2}.$

答案： 解:原式$=\left(\frac{3}{4}-\frac{5}{6}+\frac{7}{12}\right)×(-36)=\frac{3}{4}×(-36)-\frac{5}{6}×(-36)+\frac{7}{12}×(-36)$$=-27+30-21=-18$.

4. 计算:$99\frac {15}{16}×(-8).$

答案： 解:原式$=\left(100-\frac{1}{16}\right)×(-8)=100×(-8)-\frac{1}{16}×(-8)=-800+\frac{1}{2}=-799\frac{1}{2}$.

5. 计算:$(-12)×2\frac {1}{4}-\frac {9}{4}÷(-\frac {1}{8}).$

答案： 解:原式$=(-12)×\frac{9}{4}-\frac{9}{4}×(-8)=[-12-(-8)]×\frac{9}{4}=-4×\frac{9}{4}=-9$.

6. 计算:$(-\frac {1}{63})÷(\frac {1}{9}-\frac {2}{7}+\frac {2}{3}).$

答案： 解:因为$\left(\frac{1}{9}-\frac{2}{7}+\frac{2}{3}\right)÷\left(-\frac{1}{63}\right)=\left(\frac{1}{9}-\frac{2}{7}+\frac{2}{3}\right)×(-63)=\frac{1}{9}×(-63)-\frac{2}{7}×(-63)+\frac{2}{3}×(-63)=-7+18-42=-31$.所以原式$=-\frac{1}{31}$.

7. 计算:$(-2024\frac {1}{6})+2023\frac {3}{4}+(-2022\frac {5}{6})+2021\frac {1}{2}.$

答案： 解:原式$=\left[-2024+\left(-\frac{1}{6}\right)\right]+\left(2023+\frac{3}{4}\right)+\left[-2022+\left(-\frac{5}{6}\right)\right]+2021+\frac{1}{2}=[-2024+2023+(-2022)+2021]+\left[-\frac{1}{6}+\frac{3}{4}+\left(-\frac{5}{6}\right)+\frac{1}{2}\right]=-2+\frac{1}{4}=-1\frac{3}{4}$.

8. 计算:$\frac {1}{7}-\frac {1}{56}-\frac {1}{72}-\frac {1}{90}.$

答案： 解:原式$=\frac{1}{7}-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{7}-\frac{1}{7}+\frac{1}{8}-\frac{1}{8}+\frac{1}{9}-\frac{1}{9}+\frac{1}{10}=\frac{1}{10}$.