《2025年一遍过七年级数学上册沪科版》

1 计算：$45×(-25)×\frac {7}{8}×(-\frac {11}{15})÷(-\frac {1}{2})^{2}×(-1\frac {1}{7})$。

答案： 解:$45×(-25)×\frac {7}{8}×(-\frac {11}{15})÷(-\frac {1}{2})^{2}×(-1\frac {1}{7})$
$=45×(-25)×\frac {7}{8}×(-\frac {11}{15})×4×(-\frac {8}{7})$
$=-(45×\frac {11}{15})×(25×4)×(\frac {7}{8}×\frac {8}{7})$
$=-33×100×1$
$=-3300.$

2 计算：
(1)$(\frac {7}{9}-\frac {5}{6}+\frac {7}{18})×36-6×1.45+3.95×6$；
(2)$-\frac {3}{2}×(-\frac {3}{5})^{2}-2\frac {5}{19}×\frac {19}{43}×(-1\frac {1}{2})+(\frac {4}{5})^{2}×(-\frac {3}{2})$。

答案： 解:
(1)$(\frac {7}{9}-\frac {5}{6}+\frac {7}{18})×36-6×1.45+3.95×6$
$=\frac {7}{9}×36+(-\frac {5}{6})×36+\frac {7}{18}×36+6×(-1.45+3.95)$
$=28-30+14+6×2.5$
$=28-30+14+15$
$=27.$
(2)$-\frac {3}{2}×(-\frac {3}{5})^{2}-2\frac {5}{19}×\frac {19}{43}×(-1\frac {1}{2})+(\frac {4}{5})^{2}×(-\frac {3}{2})$
$=-\frac {3}{2}×[(-\frac {3}{5})^{2}-2\frac {5}{19}×\frac {19}{43}+(\frac {4}{5})^{2}]$
$=-\frac {3}{2}×(\frac {9}{25}-\frac {43}{19}×\frac {19}{43}+\frac {16}{25})$
$=-\frac {3}{2}×(\frac {9}{25}+\frac {16}{25}-1)$
$=-\frac {3}{2}×0$
$=0.$

3 计算：$99\frac {17}{18}×(-9)$。

答案： 解:$99\frac {17}{18}×(-9)$
$=(100-\frac {1}{18})×(-9)$
$=100×(-9)+(-\frac {1}{18})×(-9)$
$=-900+\frac {1}{2}$
$=-899\frac {1}{2}.$

4 [2025 安庆期中]阅读材料,回答问题。
计算：$(-\frac {1}{30})÷(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})$。
解：方法一 原式$=(-\frac {1}{30})÷[(\frac {2}{3}+\frac {1}{6})-(\frac {1}{10}+\frac {2}{5})]= (-\frac {1}{30})÷(\frac {5}{6}-\frac {1}{2})= (-\frac {1}{30})÷\frac {1}{3}= -\frac {1}{10}$。
方法二 原式的倒数为$(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})÷(-\frac {1}{30})= (\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})×(-30)= -20+3-5+12= -10$,故原式$=-\frac {1}{10}$。
根据材料用适当的方法计算：
$(-\frac {1}{42})÷(\frac {1}{6}-\frac {3}{14}+\frac {2}{3}-\frac {2}{7})$。

答案： 解:原式的倒数为$(\frac {1}{6}-\frac {3}{14}+\frac {2}{3}-\frac {2}{7})÷(-\frac {1}{42})$
$=(\frac {1}{6}-\frac {3}{14}+\frac {2}{3}-\frac {2}{7})×(-42)$
$=-7+9-28+12$
$=-14,$
故原式$=-\frac {1}{14}.$