9[2026 山东日照质检,中]用简便方法计算:
(1) $(-2024 \frac{5}{6}) + 4046 \frac{2}{3} + (-2025 \frac{2}{3}) + 1 \frac{5}{6}$.
(2) $(-199 \frac{37}{38}) × 76$.

9.【解】(1) $(-2\ 024\ \dfrac{5}{6}) + 4\ 046\ \dfrac{2}{3} + (-2\ 025\ \dfrac{2}{3}) + 1\ \dfrac{5}{6} = [ (-2\ 024) + (-\dfrac{5}{6}) ] + ( 4\ 046 + \dfrac{2}{3} ) + [ (-2\ 025) + (-\dfrac{2}{3}) ] + ( 1 + \dfrac{5}{6} ) = [ (-2\ 024) + 4\ 046 + (-2\ 025) + 1 ] + [ (-\dfrac{5}{6}) + \dfrac{2}{3} + (-\dfrac{2}{3}) + \dfrac{5}{6} ] = -2+0=-2.$
(2) $(-199\ \dfrac{37}{38}) × 76 = (-200 + \dfrac{1}{38}) × 76 = -200×76 + \dfrac{1}{38}×76 = -15\ 200+2=-15\ 198.$
技巧点拨
(1)利用拆项法,把整数结合在一起,分数结合在一起,再相加;
(2)采用拆项法把带分数拆为两项,再用乘法对加法的分配律计算.

10[2026 上海杨浦区期中,中]用简便方法计算:$\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}-\frac{21}{110}$

10.【解】原式$=\dfrac{5}{6}+\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{10}-\dfrac{1}{10}-\dfrac{1}{11} = \dfrac{5}{6}+\dfrac{1}{3}-\dfrac{1}{11} = \dfrac{77}{66}-\dfrac{6}{66}=\dfrac{71}{66}.$

11[2026安徽亳州质检,中]老师为了强化同学们的运算思维,提高数学运算能力,布置了一道有意思的计算题,鼓励大家用不同的方法计算$\frac{1}{24}÷(\frac{1}{3}-\frac{1}{4}+\frac{1}{12})$.以下是三名同学的计算过程.
甲:原式$=\frac{1}{24}÷\frac{1}{3}-\frac{1}{24}÷\frac{1}{4}+\frac{1}{24}÷\frac{1}{12}=\frac{1}{24}×3-\frac{1}{24}×4+\frac{1}{24}×12=\frac{11}{24}$.
乙:原式$=\frac{1}{24}÷(\frac{4}{12}-\frac{3}{12}+\frac{1}{12})=\frac{1}{24}÷\frac{1}{6}=\frac{1}{24}×6=\frac{1}{4}$.
丙:原式的倒数$=(\frac{1}{3}-\frac{1}{4}+\frac{1}{12})÷\frac{1}{24}=(\frac{1}{3}-\frac{1}{4}+\frac{1}{12})×24=\frac{1}{3}×24-\frac{1}{4}×24+\frac{1}{12}×24=4$,所以原式$=\frac{1}{4}$.
(1)比较他们的做法,其中的做法是错误的.
(2)选择合适的方法计算:$(-\frac{1}{210})÷(\frac{3}{7}+\frac{2}{30}-\frac{3}{10}-\frac{5}{21})$.
(3)计 算: $(\frac{3}{7}+\frac{2}{30}-\frac{3}{10}-\frac{5}{21}) ÷ (-\frac{1}{210}) × [(-\frac{1}{210})÷(\frac{3}{7}+\frac{2}{30}-\frac{3}{10}-\frac{5}{21})]$.

11.【解】(1)由甲、乙、丙的做法可知,甲的做法是错误的,故答案为甲.
(2)因为$( \dfrac{3}{7}+\dfrac{2}{30}-\dfrac{3}{10}-\dfrac{5}{21} ) ÷ (-\dfrac{1}{210}) = ( \dfrac{3}{7}+\dfrac{2}{30}-\dfrac{3}{10}-\dfrac{5}{21} ) × (-210) = \dfrac{3}{7}×(-210) + \dfrac{2}{30}×(-210) - \dfrac{3}{10}×(-210) - \dfrac{5}{21}×(-210) = -90-14+63+50=9$, 所以$(-\dfrac{1}{210}) ÷ ( \dfrac{3}{7}+\dfrac{2}{30}-\dfrac{3}{10}-\dfrac{5}{21} ) = \dfrac{1}{9}.$
(3)由(2)得原式$=9×\dfrac{1}{9}=1.$
易错警示
除法没有分配律,在进行除法运算时,将除法转化为乘法才可用乘法对加法的分配律进行运算.