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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
1.【解】原式$= 4.5+(-2.5)+9\frac{1}{3}+(-15\frac{2}{3})+2\frac{1}{3}=[ 4.5 + ( - 2.5 ) ] + [ 9 \frac{1}{3} + 2 \frac{1}{3} + ( - 15 \frac{2}{3} ) ] =2+(-4)=-2.$
2 [2025 四川自贡质检,中]计算: $-1.55×(-0.75)+(-0.55)×\frac{3}{4}$
答案
2.【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[中]计算:$(+1.75)+(-\dfrac{1}{3})+(+\dfrac{4}{5})+(+1.05)+(-\dfrac{2}{3})+(+2.2)$
答案
3.【解】原式$=(1. 75+1. 05)+(-\frac{1}{3}-\frac{2}{3})+(\frac{4}{5}+2. 2)=2. 8-1+3=4. 8.$
4 [2026 江苏常州质检, 中] $(-1.25) × \dfrac{5}{7} × (-4) × (-\dfrac{7}{5})$
答案
4.【解】原式$=[ -1. 25× (-4) ]× ( -\frac{7}{5}× \frac{5}{7} ) =5× (-1)=-5.$
5[较难]计算:$2+2-4+6-8+10-12+\dots+98-100.$
答案
5.【解】原式$=2+(2-4)+(6-8)+(10-12)+\dots+(98-100)=2+(-2)+(-2)+\dots+(-2)=2+(-2)×25=2+(-50)=-48.$
6[中]若$n=1\dfrac{1}{3}-\dfrac{7}{12}+\dfrac{9}{20}-\dfrac{11}{30}+\dfrac{13}{42}-\dfrac{15}{56}+\dfrac{17}{72}$，则$n$的负倒数是
$-\frac{9}{10}$
。答案
6.$-\frac{9}{10}$ 【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$-\frac{9}{10}$。
7[中]计算：
$(-2021 \frac{2}{7})+(-2022 \frac{4}{7})+4044+(-\frac{1}{7})$
$(-2021 \frac{2}{7})+(-2022 \frac{4}{7})+4044+(-\frac{1}{7})$
答案
解：
原式$=[(-2021)+(-\frac{2}{7})]+[(-2022)+(-\frac{4}{7})]+4044+(-\frac{1}{7})$
$=[(-2021)+(-2022)+4044]+[(-\frac{2}{7})+(-\frac{4}{7})+(-\frac{1}{7})]$
$=1+(-1)$
$=0$
原式$=[(-2021)+(-\frac{2}{7})]+[(-2022)+(-\frac{4}{7})]+4044+(-\frac{1}{7})$
$=[(-2021)+(-2022)+4044]+[(-\frac{2}{7})+(-\frac{4}{7})+(-\frac{1}{7})]$
$=1+(-1)$
$=0$
8[中]计算：
$-5 \frac{5}{6} - (+9 \frac{2}{3}) - (-17 \frac{3}{4}) + (-3 \frac{1}{2}).$
$-5 \frac{5}{6} - (+9 \frac{2}{3}) - (-17 \frac{3}{4}) + (-3 \frac{1}{2}).$
答案
解：
原式$=-5\frac{5}{6}-9\frac{2}{3}+17\frac{3}{4}-3\frac{1}{2}$
$=(-5-9+17-3)+(-\frac{5}{6}-\frac{2}{3}+\frac{3}{4}-\frac{1}{2})$
$=0+(-\frac{10}{12}-\frac{8}{12}+\frac{9}{12}-\frac{6}{12})$
$=-\frac{15}{12}$
$=-\frac{5}{4}$
原式$=-5\frac{5}{6}-9\frac{2}{3}+17\frac{3}{4}-3\frac{1}{2}$
$=(-5-9+17-3)+(-\frac{5}{6}-\frac{2}{3}+\frac{3}{4}-\frac{1}{2})$
$=0+(-\frac{10}{12}-\frac{8}{12}+\frac{9}{12}-\frac{6}{12})$
$=-\frac{15}{12}$
$=-\frac{5}{4}$
9[难]计算：$-\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}$
答案
9.【解】原式$=-(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9})=-(1-\frac{1}{9})=-\frac{8}{9}.$
10[中]计算：$(-\dfrac{1}{24})÷(\dfrac{2}{3}-\dfrac{1}{12}+\dfrac{1}{6}-\dfrac{1}{4})$
答案
10.【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$=-\frac{1}{12}.$
11 [中] $\frac{1}{2024} + \frac{2}{2024} + \frac{3}{2024} + \frac{4}{2024} + \dots + \frac{4047}{2024} = \_\_\_\_\_\_$
答案
11.$4047$ 【解析】 原式 $= (\frac{1}{2024}+\frac{4\ 047}{2\ 024}) + (\frac{2}{2\ 024}+\frac{4\ 046}{2\ 024}) + (\frac{3}{2\ 024}+\frac{4\ 045}{2\ 024}) + (\frac{4}{2\ 024}+\frac{4\ 044}{2\ 024}) + \dots + (\frac{2\ 023}{2\ 024}+\frac{2\ 025}{2\ 024}) + \frac{2\ 024}{2\ 024} = 2+2+2+2+\dots+2+1=2× 2\ 023+1=4\ 047.$
12[中]计算：$\frac{1}{2} + ( \frac{1}{3} + \frac{2}{3} ) + ( \frac{1}{4} + \frac{2}{4} + \frac{3}{4} ) + ( \frac{1}{5} + \frac{2}{5} + \frac{3}{5} + \frac{4}{5} ) + \dots + ( \frac{1}{50} + \frac{2}{50} + \frac{3}{50} + \dots + \frac{49}{50} )$
答案
12.【解】令 $S = \frac{1}{2} + (\frac{1}{3}+\frac{2}{3}) + (\frac{1}{4}+\frac{2}{4}+\frac{3}{4}) + (\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}) + \dots + (\frac{1}{50}+\frac{2}{50}+\frac{3}{50}+\dots+\frac{49}{50})$，则$S=\frac{1}{2}+(\frac{2}{3}+\frac{1}{3})+(\frac{3}{4}+\frac{2}{4}+\frac{1}{4})+(\frac{4}{5}+\frac{3}{5}+\frac{2}{5}+\frac{1}{5})+\dots+(\frac{49}{50}+\frac{48}{50}+\frac{47}{50}+\dots+\frac{1}{50})$，所以$2S=1+(1+1)+(1+1+1)+(1+1+1+1)+\dots+(1+1+1+\dots+1)=1+2+3+4+\dots+49=1\ 225,$所以 $S=612. 5,$即原式$=612. 5.$
