《2025年一遍过七年级数学上册苏科版》

1 计算：
(1)$-\frac {2}{3}-\frac {3}{5}+5+\frac {2}{3}-\frac {2}{5}+4;$
(2)$2\frac {7}{8}+(-2\frac {7}{12})+5\frac {3}{5}+(-1\frac {7}{8})+2\frac {2}{5}+(-3\frac {5}{12}).$

答案： 1 解:
(1)$-\frac {2}{3}-\frac {3}{5}+5+\frac {2}{3}-\frac {2}{5}+4$
$=(-\frac {2}{3}+\frac {2}{3})+(-\frac {3}{5}-\frac {2}{5})+(5+4)$
$=0-1+9$
$=8.$
(2)$2\frac {7}{8}+(-2\frac {7}{12})+5\frac {3}{5}+(-1\frac {7}{8})+2\frac {2}{5}+(-3\frac {5}{12})$
$=[2\frac {7}{8}+(-1\frac {7}{8})]+[(-2\frac {7}{12})+(-3\frac {5}{12})]+(5\frac {3}{5}+2\frac {2}{5})$
$=1+(-6)+8$
$=3.$

2 计算：
(1)$(-5)÷(-1\frac {2}{7})×\frac {4}{5}×(-2\frac {1}{4})÷1\frac {3}{4};$
(2)$45×(-25)×\frac {7}{8}×(-\frac {11}{15})÷\frac {1}{4}×(-1\frac {1}{7}).$

答案： 2 解:
(1)$(-5)÷(-1\frac {2}{7})×\frac {4}{5}×(-2\frac {1}{4})÷1\frac {3}{4}$
$=(-5)×(-\frac {7}{9})×\frac {4}{5}×(-\frac {9}{4})×\frac {4}{7}$
$=-(5×\frac {4}{5})×(\frac {7}{9}×\frac {9}{4}×\frac {4}{7})$
$=-4×1$
$=-4.$
(2)$45×(-25)×\frac {7}{8}×(-\frac {11}{15})÷\frac {1}{4}×(-1\frac {1}{7})$
$=-45×25×\frac {7}{8}×\frac {11}{15}×4×\frac {8}{7}$
$=-(45×\frac {11}{15})×(25×4)×(\frac {7}{8}×\frac {8}{7})$
$=-33×100×1$
$=-3300.$

3 计算：
(1)$(-8)×(-\frac {1}{6}-\frac {5}{12}+\frac {3}{10})×15;$
(2)一题多解$57×\frac {55}{56}+27×\frac {27}{28};$
(3)$(-375)×(-8)+(-375)×(-9)+375×(-7).$

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)通解$57×\frac {55}{56}+27×\frac {27}{28}=(56+1)×\frac {55}{56}+(28-1)×\frac {27}{28}=56×\frac {55}{56}+1×\frac {55}{56}+28×\frac {27}{28}-1×\frac {27}{28}=55+\frac {55}{56}+27-\frac {27}{28}=82+\frac {1}{56}=82\frac {1}{56}.$
巧解$57×\frac {55}{56}+27×\frac {27}{28}=57×(1-\frac {1}{56})+27×(1-\frac {1}{28})=57-\frac {57}{56}+27-\frac {27}{28}=(57+27)-(\frac {57}{56}+\frac {27}{28})=84-1\frac {55}{56}=82\frac {1}{56}.$
(3)$(-375)×(-8)+(-375)×(-9)+375×(-7)=375×(8+9-7)=375×10=3750.$

4 计算：$\frac {1}{1×3}+\frac {1}{3×5}+\frac {1}{5×7}+... +\frac {1}{2023×2025}.$

答案： 4 解:$\frac {1}{1×3}+\frac {1}{3×5}+\frac {1}{5×7}+... +\frac {1}{2023×2025}=\frac {1}{2}×(1-\frac {1}{3})+\frac {1}{2}×(\frac {1}{3}-\frac {1}{5})+\frac {1}{2}×(\frac {1}{5}-\frac {1}{7})+... +\frac {1}{2}×(\frac {1}{2023}-\frac {1}{2025})=\frac {1}{2}×(1-\frac {1}{2025})=\frac {1012}{2025}.$

(1)把下列各式写成去掉绝对值符号的形式(不用写出计算结果):
①$|\frac {2}{3}-\frac {2}{5}|=$$\frac {2}{3}-\frac {2}{5}$;②$|3.14-π|=$$π-3.14$;
(2)当$a＞b$时,$|a-b|=$$a-b$;当$a\lt b$时,$|a-b|=$$b-a$;
(3)计算:$|\frac {1}{2}-1|+|\frac {1}{3}-\frac {1}{2}|+|\frac {1}{4}-\frac {1}{3}|+... +|\frac {1}{2025}-\frac {1}{2024}|.$
$|\frac {1}{2}-1|+|\frac {1}{3}-\frac {1}{2}|+|\frac {1}{4}-\frac {1}{3}|+... +|\frac {1}{2025}-\frac {1}{2024}|=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+... +\frac {1}{2024}-\frac {1}{2025}=1-\frac {1}{2025}=\frac {2024}{2025}.$

答案： 5 解:
(1)①$\frac {2}{3}-\frac {2}{5}$;②$π-3.14$
(2)$a-b$ $b-a$
(3)$|\frac {1}{2}-1|+|\frac {1}{3}-\frac {1}{2}|+|\frac {1}{4}-\frac {1}{3}|+... +|\frac {1}{2025}-\frac {1}{2024}|=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+... +\frac {1}{2024}-\frac {1}{2025}=1-\frac {1}{2025}=\frac {2024}{2025}.$