《2025年初中必刷题七年级数学上册北师大版》

1 [2025四川泸州期末,中]计算:$4.5+[(-2.5)+9\frac {1}{3}+(-15\frac {2}{3})]+2\frac {1}{3}.$

答案： 【解】原式$=4.5+(-2.5)+9\frac{1}{3}+(-15\frac{2}{3})+2\frac{1}{3}=[4.5+(-2.5)]+[9\frac{1}{3}+(-15\frac{2}{3})+2\frac{1}{3}]=2+(-4)=-2.$

2 [2025四川自贡质检,中]计算:$-1.55×(-0.75)+(-0.55)×\frac {3}{4}.$

答案： 【解】原式$=1.55×\frac{3}{4}+(-0.55)×\frac{3}{4}=(1.55-0.55)×\frac{3}{4}=1×\frac{3}{4}=\frac{3}{4}.$

3 [2024浙江舟山调研,中]计算:
(1)$(+1.75)+(-\frac {1}{3})+(+\frac {4}{5})+(+1.05)+(-\frac {2}{3})+(+2.2);$
(2)$-4.27+3.8-0.73+1.2.$

答案： 【解】
(1)原式$=(1.75+1.05)+(-\frac{1}{3}-\frac{2}{3})+(\frac{4}{5}+2.2)=2.8-1+3=4.8.(2)$原式=(-4.27-0.73)+(3.8+1.2)=-5+5=0.

4 [较难]计算:$2+2-4+6-8+10-12+... +98-100.$

答案： 【解】原式=2+(2-4)+(6-8)+(10-12)+…+(98-100)=2+(-2)+(-2)+…+(-2)=2+(-2)×25=2+(-50)=-48.

5 [2024河南平顶山调研,中]若$n=1\frac {1}{3}-\frac {7}{12}+\frac {9}{20}-\frac {11}{30}+\frac {13}{42}-\frac {15}{56}+\frac {17}{72}$,则n的负倒数是____.

答案： $-\frac{9}{10}【$解析】因为$n=1\frac{1}{3}-\frac{7}{12}+\frac{9}{20}-\frac{11}{30}+\frac{13}{42}-\frac{15}{56}+\frac{17}{72}=1+\frac{1}{3}-(\frac{1}{3}+\frac{1}{4})+(\frac{1}{4}+\frac{1}{5})-(\frac{1}{5}+\frac{1}{6})+(\frac{1}{6}+\frac{1}{7})-(\frac{1}{7}+\frac{1}{8})+(\frac{1}{8}+\frac{1}{9})=1+\frac{1}{3}-\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+\frac{1}{5}-\frac{1}{5}-\frac{1}{6}+\frac{1}{6}+\frac{1}{7}-\frac{1}{7}-\frac{1}{8}+\frac{1}{8}+\frac{1}{9}=1+\frac{1}{9}=\frac{10}{9},$所以n的负倒数是$-\frac{9}{10}.$

6 [中]计算:$(-2021\frac {2}{7})+(-2022\frac {4}{7})+4044+(-\frac {1}{7}).$

答案： 【解】原式$=(-2021\frac{2}{7})+(-2022\frac{4}{7})+4044-\frac{1}{7}=-2021-\frac{2}{7}-2022-\frac{4}{7}+4044-\frac{1}{7}=(-2021-2022+4044)+(-\frac{2}{7}-\frac{4}{7}-\frac{1}{7})=1+(-1)=0.$

7 [2024北京平谷区期中,中]计算:$-5\frac {5}{6}-(+9\frac {2}{3})-(-17\frac {3}{4})+(-3\frac {1}{2}).$

答案： 【解】原式$=-5\frac{5}{6}+(-9\frac{2}{3})+(+17\frac{3}{4})+(-3\frac{1}{2})=[(-5)+(-9)+17+(-3)]+[(-\frac{5}{6})+(-\frac{2}{3})+\frac{3}{4}+(-\frac{1}{2})]=0+(-1\frac{1}{4})=-1\frac{1}{4}.$

8 [难]计算:$-\frac {1}{2}-\frac {1}{6}-\frac {1}{12}-\frac {1}{20}-\frac {1}{30}-\frac {1}{42}-\frac {1}{56}-\frac {1}{72}.$

答案： 【解】原式$=-(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72})=-[(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{9})]=-(1-\frac{1}{9})=-\frac{8}{9}.$

9 [中]计算:$(-\frac {1}{24})÷(\frac {2}{3}-\frac {1}{12}+\frac {1}{6}-\frac {1}{4}).$

答案： 【解】原式的倒数为$(\frac{2}{3}-\frac{1}{12}+\frac{1}{6}-\frac{1}{4})÷(-\frac{1}{24})=(\frac{2}{3}-\frac{1}{12}+\frac{1}{6}-\frac{1}{4})×(-24)=\frac{2}{3}×(-24)-\frac{1}{12}×(-24)+\frac{1}{6}×(-24)-\frac{1}{4}×(-24)=-16+2-4+6=-12,$所以原式$=-\frac{1}{12}.$

10 [中]$\frac {1}{2024}+\frac {2}{2024}+\frac {3}{2024}+\frac {4}{2024}+... +\frac {4047}{2024}=$____.

答案： 4047【解析】原式$=(\frac{1}{2024}+\frac{4047}{2024})+(\frac{2}{2024}+\frac{4046}{2024})+(\frac{3}{2024}+\frac{4045}{2024})+(\frac{4}{2024}+\frac{4044}{2024})+…+(\frac{2023}{2024}+\frac{2025}{2024})+\frac{2024}{2024}=2+2+2+2+…+2+1=2×2023+1=4047.$

11 [中]计算:$\frac {1}{2}+(\frac {1}{3}+\frac {2}{3})+(\frac {1}{4}+\frac {2}{4}+\frac {3}{4})+(\frac {1}{5}+\frac {2}{5}+\frac {3}{5}+\frac {4}{5})+... +(\frac {1}{50}+\frac {2}{50}+\frac {3}{50}+... +\frac {49}{50}).$

答案： 【解】令$S=\frac{1}{2}+(\frac{1}{3}+\frac{2}{3})+(\frac{1}{4}+\frac{2}{4}+\frac{3}{4})+(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5})+…+(\frac{1}{50}+\frac{2}{50}+\frac{3}{50}+…+\frac{49}{50}),$则$S=\frac{1}{2}+(\frac{2}{3}+\frac{1}{3})+(\frac{3}{4}+\frac{2}{4}+\frac{1}{4})+(\frac{4}{5}+\frac{3}{5}+\frac{2}{5}+\frac{1}{5})+…+(\frac{49}{50}+\frac{48}{50}+\frac{47}{50}+…+\frac{1}{50}),$即2S=1+(1+1)+(1+1+1)+(1+1+1+1)+…+(1+1+1+…+1)=1+2+3+4+…+49=1225,所以S=612.5,即原式=612.5.