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4. 计算：
(1) $2\frac{1}{3} - (+10\frac{1}{3}) + (-8\frac{1}{5}) - 3\frac{2}{5}$
(2) $(-17\frac{2}{3}) + 16\frac{3}{4} + (-15\frac{1}{3}) - 2\frac{1}{2}$
(3) $-8721 + 53\frac{19}{21} - 1279 - 43\frac{19}{21}$
(4) $-5\frac{5}{6} + (-9\frac{2}{3}) + 17\frac{3}{4} + (-3\frac{1}{2})$
(1) $2\frac{1}{3} - (+10\frac{1}{3}) + (-8\frac{1}{5}) - 3\frac{2}{5}$
(2) $(-17\frac{2}{3}) + 16\frac{3}{4} + (-15\frac{1}{3}) - 2\frac{1}{2}$
(3) $-8721 + 53\frac{19}{21} - 1279 - 43\frac{19}{21}$
(4) $-5\frac{5}{6} + (-9\frac{2}{3}) + 17\frac{3}{4} + (-3\frac{1}{2})$
答案
(1) 原式$=2+\dfrac{1}{3}-10-\dfrac{1}{3}-8-\dfrac{1}{5}-3-\dfrac{2}{5}=(2-10-8-3)+(\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{5})=-19-\dfrac{3}{5}=-19\dfrac{3}{5}$.
(2) 原式$=[(-17)+16+(-15)+(-2)]+[(-\dfrac{2}{3})+\dfrac{3}{4}+(-\dfrac{1}{3})+(-\dfrac{1}{2})]=-18+(-\dfrac{3}{4})=-18\dfrac{3}{4}$.
(3) 原式$=-8\ 721+53+\dfrac{19}{21}-1\ 279-43-\dfrac{19}{21}=(-8\ 721+53-1\ 279-43)+(\dfrac{19}{21}-\dfrac{19}{21})=-9\ 990$.
(4) 原式$=[(-5)+(-\dfrac{5}{6})]+[(-9)+(-\dfrac{2}{3})]+(17+\dfrac{3}{4})+[(-3)+(-\dfrac{1}{2})]=[(-5)+(-9)+17+(-3)]+[(-\dfrac{5}{6})+(-\dfrac{2}{3})+\dfrac{3}{4}+(-\dfrac{1}{2})]=0+(-1\dfrac{1}{4})=-1\dfrac{1}{4}$.
(2) 原式$=[(-17)+16+(-15)+(-2)]+[(-\dfrac{2}{3})+\dfrac{3}{4}+(-\dfrac{1}{3})+(-\dfrac{1}{2})]=-18+(-\dfrac{3}{4})=-18\dfrac{3}{4}$.
(3) 原式$=-8\ 721+53+\dfrac{19}{21}-1\ 279-43-\dfrac{19}{21}=(-8\ 721+53-1\ 279-43)+(\dfrac{19}{21}-\dfrac{19}{21})=-9\ 990$.
(4) 原式$=[(-5)+(-\dfrac{5}{6})]+[(-9)+(-\dfrac{2}{3})]+(17+\dfrac{3}{4})+[(-3)+(-\dfrac{1}{2})]=[(-5)+(-9)+17+(-3)]+[(-\dfrac{5}{6})+(-\dfrac{2}{3})+\dfrac{3}{4}+(-\dfrac{1}{2})]=0+(-1\dfrac{1}{4})=-1\dfrac{1}{4}$.
5. 计算：
(1) $\dfrac{1}{30} ÷ ( \dfrac{2}{3} - \dfrac{1}{10} + \dfrac{1}{6} - \dfrac{2}{5} ).$
(2) $( -\dfrac{1}{56} ) ÷ ( \dfrac{1}{7} + \dfrac{3}{28} - \dfrac{1}{4} + \dfrac{5}{8} ).$
(1) $\dfrac{1}{30} ÷ ( \dfrac{2}{3} - \dfrac{1}{10} + \dfrac{1}{6} - \dfrac{2}{5} ).$
(2) $( -\dfrac{1}{56} ) ÷ ( \dfrac{1}{7} + \dfrac{3}{28} - \dfrac{1}{4} + \dfrac{5}{8} ).$
答案
(1) 原式的倒数$=(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})÷\dfrac{1}{30}=(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})×30=\dfrac{2}{3}×30-\dfrac{1}{10}×30+\dfrac{1}{6}×30-\dfrac{2}{5}×30=20-3+5-12=10$. 所以原式$=\dfrac{1}{10}$.
(2) 原式的倒数$=(\dfrac{1}{7}+\dfrac{3}{28}-\dfrac{1}{4}+\dfrac{5}{8})÷(-\dfrac{1}{56})=(\dfrac{1}{7}+\dfrac{3}{28}-\dfrac{1}{4}+\dfrac{5}{8})×(-56)=\dfrac{1}{7}×(-56)+\dfrac{3}{28}×(-56)-\dfrac{1}{4}×(-56)+\dfrac{5}{8}×(-56)=-8-6+14-35=-35$. 所以原式$=-\dfrac{1}{35}$.
(2) 原式的倒数$=(\dfrac{1}{7}+\dfrac{3}{28}-\dfrac{1}{4}+\dfrac{5}{8})÷(-\dfrac{1}{56})=(\dfrac{1}{7}+\dfrac{3}{28}-\dfrac{1}{4}+\dfrac{5}{8})×(-56)=\dfrac{1}{7}×(-56)+\dfrac{3}{28}×(-56)-\dfrac{1}{4}×(-56)+\dfrac{5}{8}×(-56)=-8-6+14-35=-35$. 所以原式$=-\dfrac{1}{35}$.
6. 计算：
(1) $\frac{2}{5} - \left|-1\frac{1}{2}\right| - (+2\frac{1}{4}) - (-2.75)$。
(2) $-32\frac{1}{3} - [5\frac{1}{4} - (+3\frac{1}{7}) - 3.25 - 2\frac{6}{7}]$。
(3) $\left|\frac{1}{3} - 1\right| + \left|\frac{1}{5} - \frac{1}{3}\right| + \left|\frac{1}{7} - \frac{1}{5}\right| + \dots + \left|\frac{1}{97} - \frac{1}{95}\right| + \left|\frac{1}{99} - \frac{1}{97}\right|$。
(1) $\frac{2}{5} - \left|-1\frac{1}{2}\right| - (+2\frac{1}{4}) - (-2.75)$。
(2) $-32\frac{1}{3} - [5\frac{1}{4} - (+3\frac{1}{7}) - 3.25 - 2\frac{6}{7}]$。
(3) $\left|\frac{1}{3} - 1\right| + \left|\frac{1}{5} - \frac{1}{3}\right| + \left|\frac{1}{7} - \frac{1}{5}\right| + \dots + \left|\frac{1}{97} - \frac{1}{95}\right| + \left|\frac{1}{99} - \frac{1}{97}\right|$。
答案
(1) 原式$=\dfrac{2}{5}-1\dfrac{1}{2}-2\dfrac{1}{4}+2.75=\dfrac{2}{5}-1-\dfrac{1}{2}+(2.75-2\dfrac{1}{4})=\dfrac{2}{5}-1-\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{2}{5}-1+(-\dfrac{1}{2}+\dfrac{1}{2})=\dfrac{2}{5}-1=-\dfrac{3}{5}$.
(2) 原式$=-32\dfrac{1}{3}-[(5\dfrac{1}{4}-3.25)+(-3\dfrac{1}{7}-2\dfrac{6}{7})]=-32\dfrac{1}{3}-[2+(-6)]=-32\dfrac{1}{3}-(-4)=-32\dfrac{1}{3}+4=-28\dfrac{1}{3}$.
(3) 原式$=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dots+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}=1+(-\dfrac{1}{3}+\dfrac{1}{3})+(-\dfrac{1}{5}+\dfrac{1}{5})+(-\dfrac{1}{7}+\dfrac{1}{7})+\dots+(-\dfrac{1}{95}+\dfrac{1}{95})+(-\dfrac{1}{97}+\dfrac{1}{97})-\dfrac{1}{99}=1-\dfrac{1}{99}=\dfrac{98}{99}$.
(2) 原式$=-32\dfrac{1}{3}-[(5\dfrac{1}{4}-3.25)+(-3\dfrac{1}{7}-2\dfrac{6}{7})]=-32\dfrac{1}{3}-[2+(-6)]=-32\dfrac{1}{3}-(-4)=-32\dfrac{1}{3}+4=-28\dfrac{1}{3}$.
(3) 原式$=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dots+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}=1+(-\dfrac{1}{3}+\dfrac{1}{3})+(-\dfrac{1}{5}+\dfrac{1}{5})+(-\dfrac{1}{7}+\dfrac{1}{7})+\dots+(-\dfrac{1}{95}+\dfrac{1}{95})+(-\dfrac{1}{97}+\dfrac{1}{97})-\dfrac{1}{99}=1-\dfrac{1}{99}=\dfrac{98}{99}$.
7. 计算 $1 - 5 + 5^2 - 5^3 + 5^4 - 5^5 + \dots + 5^{2020} - 5^{2021} + 5^{2022} - \dfrac{5^{2023}}{6}$ 的结果为
$\dfrac{1}{6}$
。答案
令$S=1-5+5^2-5^3+5^4-5^5+\dots+5^{2020}-5^{2021}+5^{2022}$, 则$5S=5-5^2+5^3-5^4+5^5-5^6+\dots+5^{2021}-5^{2022}+5^{2023}$. 所以$5S+S=1+5^{2023}$. 所以$S=\dfrac{1+5^{2023}}{6}$. 所以$1-5+5^2-5^3+5^4-5^5+\dots+5^{2020}-5^{2021}+5^{2022}-\dfrac{5^{2023}}{6}=\dfrac{1+5^{2023}}{6}-\dfrac{5^{2023}}{6}=\dfrac{1}{6}$.
