《2025年一本计算题满分训练七年级数学全一册人教版》

1.【一题多解】$(-7\frac {1}{2})+(+2\frac {2}{5})+(-\frac {1}{4}).$

答案： 一题多解 解:方法1:
原式$=-7+(-\frac {1}{2})+2+\frac {2}{5}+(-\frac {1}{4})$
$=-(7-2)+[(-\frac {1}{2})+(-\frac {1}{4})+\frac {2}{5}]$
$=-5+[-( \frac {1}{2}+\frac {1}{4})+\frac {2}{5}]$
$=-5+(-\frac {3}{4}+\frac {2}{5})$
$=-5+[-( \frac {3}{4}-\frac {2}{5})]$
$=-5+(-\frac {7}{20})$
$=-(5+\frac {7}{20})$
$=-5\frac {7}{20}.$
方法2:
原式$=(-7\frac {1}{2})+(-\frac {1}{4})+(+2\frac {2}{5})$
$=-(7\frac {1}{2}+\frac {1}{4})+2\frac {2}{5}=-7\frac {3}{4}+2\frac {2}{5}$
$=-(7\frac {3}{4}-2\frac {2}{5})=-5\frac {7}{20}.$

2.$(+4\frac {5}{6})+(-3\frac {2}{3})+(-\frac {1}{6}).$

答案： 解:原式$=4+\frac {5}{6}+(-3)+(-\frac {2}{3})+(-\frac {1}{6})$
$=(4-3)+(\frac {5}{6}-\frac {1}{6}-\frac {2}{3})=1.$

3.$6\frac {4}{9}+(-2\frac {1}{7})+(-2\frac {2}{3})+\frac {15}{7}.$

答案： 解:原式$=6+\frac {4}{9}+(-2)+(-\frac {1}{7})+(-2)+$
$(-\frac {2}{3})+2+\frac {1}{7}=(6-2-2+2)+(\frac {4}{9}-$
$\frac {1}{7}-\frac {2}{3}+\frac {1}{7})=4+(-\frac {2}{9})=+(4-\frac {2}{9})=3\frac {7}{9}.$

4.$1\frac {2}{3}+(-1\frac {2}{5})+\frac {4}{3}+(-1)+(-3\frac {3}{5}).$

答案： 解:原式$=(1\frac {2}{3}+\frac {4}{3})+(-1)+[(-1\frac {2}{5})+$
$(-3\frac {3}{5})]=3+(-1)+(-5)=-3.$

5.【一题多解】$(-9\frac {5}{12})+15\frac {3}{4}+(-3\frac {1}{4})+(-22\frac {1}{2})+(-15\frac {7}{12}).$

答案： 一题多解 解:方法1:同号结合法
原式$=[(-9\frac {5}{12})+(-15\frac {7}{12})]+15\frac {3}{4}+$
$[(-3\frac {1}{4})+(-22\frac {1}{2})]=(-25)+15\frac {3}{4}+$
$(-25\frac {3}{4})$
$=[(-25)+(-25\frac {3}{4})]+15\frac {3}{4}=(-50\frac {3}{4})+15\frac {3}{4}$
$=-35.$
方法2:同分母结合法
原式$=[(-9\frac {5}{12})+(-15\frac {7}{12})]+15\frac {3}{4}+$
$[(-3\frac {1}{4})+(-22\frac {1}{2})]=(-25)+15\frac {3}{4}+$
$(-25\frac {3}{4})$
$=(-25)+[15\frac {3}{4}+(-25\frac {3}{4})]=(-25)+(-10)$
$=-35.$

6.$(-18\frac {4}{5})+(+53\frac {3}{5})+(-53\frac {3}{5})+(+18\frac {4}{5})+(-100).$

答案： 解:原式$=[(-18\frac {4}{5})+(+18\frac {4}{5})]+[(+53\frac {3}{5})+$
$(-53\frac {3}{5})]+(-100)=-100.$

7.$(-11\frac {1}{2})+(+23\frac {1}{3})+(+21\frac {1}{2})+(-34\frac {1}{3})+(-27).$

答案： 解:原式$=[(-11\frac {1}{2})+(+21\frac {1}{2})]+[(+23\frac {1}{3})+(-34\frac {1}{3})]+(-27)=10+(-11)+(-27)=-28.$