三、解答题（共 52 分）
13.（20 分）计算：
(1)$(-3\dfrac{1}{5})×(-\dfrac{2}{7})÷(+1\dfrac{3}{5})$；
(2)$(-\dfrac{9}{16})÷(-\dfrac{3}{8})÷(-\dfrac{3}{2})$；
(3)$(-81)×(-\dfrac{4}{9})÷\dfrac{9}{4}×\dfrac{1}{8}$；
(4)$(-2)÷(-\dfrac{3}{7})×\dfrac{4}{7}÷(-5\dfrac{1}{7})$。

13. (1)原式=$\dfrac{16}{5}×\dfrac{2}{7}×\dfrac{5}{8}=\dfrac{4}{7}$。
(2)原式=$-\dfrac{9}{16}×\dfrac{8}{3}×\dfrac{2}{3}=-1$。
(3)原式=$81×\dfrac{4}{9}×\dfrac{4}{9}×\dfrac{1}{8}=2$。
(4)原式=$-2×\dfrac{7}{3}×\dfrac{4}{7}×\dfrac{7}{36}=-\dfrac{14}{27}$。

14. (20 分)计算：
(1)$(-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5})×(-20)$；
(2)$-\dfrac{5}{6}×(12-2\dfrac{2}{5}-0.6)$；
(3)$49\dfrac{24}{25}×(-5)$；
(4)$(-36\dfrac{4}{5})×\dfrac{1}{6}$；

14. (1)原式=$-\dfrac{1}{2}×(-20)+\dfrac{1}{3}×(-20)-\dfrac{1}{4}×(-20)-\dfrac{1}{5}×(-20)=10-\dfrac{20}{3}+5+4=19-\dfrac{20}{3}=\dfrac{37}{3}$。
(2)原式=$(-\dfrac{5}{6})×12-\dfrac{12}{5}×(-\dfrac{5}{6})-\dfrac{3}{5}×(-\dfrac{5}{6})=-10+2+\dfrac{1}{2}=-8+\dfrac{1}{2}=-7\dfrac{1}{2}$。
(3)原式=$(50-\dfrac{1}{25})×(-5)=50×(-5)-\dfrac{1}{25}×(-5)=-250+\dfrac{1}{5}=-249\dfrac{4}{5}$。
(4)原式=$(-36-\dfrac{4}{5})×\dfrac{1}{6}=-36×\dfrac{1}{6}-\dfrac{4}{5}×\dfrac{1}{6}=-6-\dfrac{2}{15}=-6\dfrac{2}{15}$。
(5)原式=$-13×\dfrac{2}{3}+\dfrac{1}{3}×(-13)-0.34×\dfrac{2}{7}-\dfrac{5}{7}×0.34=-13×(\dfrac{2}{3}+\dfrac{1}{3})-0.34×(\dfrac{2}{7}+\dfrac{5}{7})=-13×1-0.34×1=-13-0.34=-13.34$。

15. (12 分)根据所学知识解答下列问题.
(1) 根据倒数的定义我们知道, 若 $(a+b) ÷ c=3$, 则 $c ÷(a+b)=$.
(2) 计算: $(-\dfrac{2}{9}+\dfrac{1}{4}-\dfrac{5}{12}) ÷(-\dfrac{1}{36})$.
(3) 根据以上信息可知, $(-\dfrac{1}{36}) ÷(-\dfrac{2}{9}+\dfrac{1}{4}-\dfrac{5}{12})=$.

15. (1)$\dfrac{1}{3}$
(2) $(-\dfrac{2}{9}+\dfrac{1}{4}-\dfrac{5}{12}) ÷(-\dfrac{1}{36})=(-\dfrac{2}{9}+\dfrac{1}{4}-\dfrac{5}{12}) ×(-36)=-\dfrac{2}{9}×(-36)+\dfrac{1}{4}×(-36)-\dfrac{5}{12}×(-36)=8-9+15=14$。
(3)$\dfrac{1}{14}$ 解析:由(2)可知,$(-\dfrac{2}{9}+\dfrac{1}{4}-\dfrac{5}{12}) ÷(-\dfrac{1}{36})=14$, 所以$(-\dfrac{1}{36}) ÷(-\dfrac{2}{9}+\dfrac{1}{4}-\dfrac{5}{12})=\dfrac{1}{14}$。