《2025年一遍过七年级数学上册华师大版》

1 计算:
(1) $ -\dfrac{2}{7}-(-\dfrac{4}{9})+\dfrac{4}{7}-\dfrac{2}{9}-\dfrac{1}{7} $;
(2) $ (-3\dfrac{4}{7})+12.5+(-16\dfrac{3}{7})-(-2.5) $.
(3) $ (-5\dfrac{5}{6})+(-9\dfrac{2}{3})+17\dfrac{3}{4}+(-3\dfrac{1}{2}) $

答案： 1 解:
(1)$-\frac {2}{7}-(-\frac {4}{9})+\frac {4}{7}-\frac {2}{9}-\frac {1}{7}$
$=-\frac {2}{7}+\frac {4}{9}+\frac {4}{7}-\frac {2}{9}-\frac {1}{7}$
$=(-\frac {2}{7}+\frac {4}{7}-\frac {1}{7})+(\frac {4}{9}-\frac {2}{9})$
$=\frac {1}{7}+\frac {2}{9}$
$=\frac {23}{63}.$
(2)$(-3\frac {4}{7})+12.5+(-16\frac {3}{7})-(-2.5)$
$=(-3\frac {4}{7}-16\frac {3}{7})+(12.5+2.5)$
$=-20+15$
$=-5.$
(3)$(-5\frac {5}{6})+(-9\frac {2}{3})+17\frac {3}{4}+(-3\frac {1}{2})$
$=[(-5)+(-\frac {5}{6})]+[(-9)+(-\frac {2}{3})]+(17+\frac {3}{4})+$
$[(-3)+(-\frac {1}{2})]$
$=[(-5)+(-9)+17+(-3)]+[(-\frac {5}{6})+(-\frac {2}{3})+\frac {3}{4}+$
$(-\frac {1}{2})]$
$=0+(-1\frac {1}{4})$
$=-1\frac {1}{4}.$

2 计算:
$ 45×(-25)×\dfrac{7}{8}×(-\dfrac{11}{15})÷\dfrac{1}{4}×(-1\dfrac{1}{7}) $.

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

3 计算:
(1) $ (-8)×(-\dfrac{1}{6}-\dfrac{5}{12}+\dfrac{3}{10})×15 $;
(2) $ -\dfrac{3}{2}×\dfrac{9}{25}-2\dfrac{5}{19}×\dfrac{19}{43}×(-1\dfrac{1}{2})+\dfrac{16}{25}×(-\dfrac{3}{2}) $.

答案： 3 解:
(1)$(-8)×(-\frac {1}{6}-\frac {5}{12}+\frac {3}{10})×15$
$=(-8)×15×(-\frac {1}{6}-\frac {5}{12}+\frac {3}{10})$
$=(-120)×(-\frac {1}{6}-\frac {5}{12}+\frac {3}{10})$
$=(-120)×(-\frac {1}{6})+(-120)×(-\frac {5}{12})+(-120)×\frac {3}{10}$
$=20+50-36$
$=34.$
(2)$-\frac {3}{2}×\frac {9}{25}-2\frac {5}{19}×\frac {19}{43}×(-1\dfrac{1}{2})+\dfrac{16}{25}×(-\dfrac{3}{2})$
$=-\frac {3}{2}×(\frac {9}{25}-2\frac {5}{19}×\frac {19}{43}+\frac {16}{25})$
$=-\frac {3}{2}×(\frac {9}{25}-\frac {43}{19}×\frac {19}{43}+\frac {16}{25})$
$=-\frac {3}{2}×(\frac {9}{25}+\frac {16}{25}-1)$
$=-\frac {3}{2}×0$
$=0.$

4 计算 $ (-\dfrac{1}{30})÷(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5}) $时,按照运算顺序计算比较麻烦,可以采用倒数法:
因为原式的倒数为 $ (\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})÷(-\dfrac{1}{30})= (\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})×(-30)= -20+3-5+12= -10 $,所以原式 $ =-\dfrac{1}{10} $.
请用上述方法计算 $ (-\dfrac{1}{56})÷(\dfrac{3}{8}-\dfrac{3}{14}+\dfrac{1}{2}-\dfrac{2}{7}) $.

答案： 4 解:$(-\frac {1}{56})÷(\frac {3}{8}-\frac {3}{14}+\frac {1}{2}-\frac {2}{7})$的倒数为$(\frac {3}{8}-\frac {3}{14}+$
$\frac {1}{2}-\frac {2}{7})÷(-\frac {1}{56}),$
$(\frac {3}{8}-\frac {3}{14}+\frac {1}{2}-\frac {2}{7})÷(-\frac {1}{56})$
$=(\frac {3}{8}-\frac {3}{14}+\frac {1}{2}-\frac {2}{7})×(-56)$
$=-\frac {3}{8}×56+\frac {3}{14}×56-\frac {1}{2}×56+\frac {2}{7}×56$
$=-21+12-28+16=-21,$
所以原式$=-\frac {1}{21}.$

5 计算: $ (-\dfrac{1}{2})+(\dfrac{1}{3}+\dfrac{2}{3})+(-\dfrac{1}{4}-\dfrac{2}{4}-\dfrac{3}{4})+(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5})+…+(\dfrac{1}{55}+\dfrac{2}{55}+…+\dfrac{53}{55}+\dfrac{54}{55}) $.

答案： 5 解:$(-\frac {1}{2})+(\frac {1}{3}+\frac {2}{3})+(-\frac {1}{4}-\frac {2}{4}-\frac {3}{4})+(\frac {1}{5}+\frac {2}{5}+$
$\frac {3}{5}+\frac {4}{5})+... +(\frac {1}{55}+\frac {2}{55}+... +\frac {53}{55}+\frac {54}{55})$
$=-\frac {1}{2}+1-\frac {6}{4}+\frac {10}{5}+... +\frac {1}{55}×\frac {(1+54)×54}{2}$
$=-\frac {1}{2}+1-\frac {3}{2}+2-\frac {5}{2}+3-\frac {7}{2}+4+... -\frac {53}{2}+27$
$=(-\frac {1}{2}+1)+(-\frac {3}{2}+2)+(-\frac {5}{2}+3)+(-\frac {7}{2}+4)+... +$
$(-\frac {53}{2}+27)$
$=\frac {1}{2}+\frac {1}{2}+\frac {1}{2}+\frac {1}{2}+... +\frac {1}{2}$(点拨:共有27个$\frac {1}{2})$
$=\frac {27}{2}.$