3. 计算:
(1) $91\frac{3}{4} × (-8)$;
(2) $(-19\frac{17}{18}) × (-18)$;
(3) $14\frac{4}{7} × (-\frac{5}{17})$.

(1)原式$=(90+\dfrac{7}{4})×(-8)=90×(-8)+\dfrac{7}{4}×(-8)=-720-14=-734.$
(2)原式$=(20-\dfrac{1}{18})×18=20×18-\dfrac{1}{18}×18=360-1=359.$
(3)原式$=(17-2\dfrac{3}{7})×(-\dfrac{5}{17})=17×(-\dfrac{5}{17})-\dfrac{17}{7}×(-\dfrac{5}{17})=-5+\dfrac{5}{7}=-4\dfrac{2}{7}.$

类型4 将被除数和除数颠倒位置
4.阅读下列材料：
计算：$\frac{1}{12} ÷ ( \frac{1}{3} - \frac{1}{4} + \frac{1}{12} )$。
解：原式的倒数为
$( \frac{1}{3} - \frac{1}{4} + \frac{1}{12} ) ÷ \frac{1}{12}$
$= ( \frac{1}{3} - \frac{1}{4} + \frac{1}{12} ) × 12$
$= \frac{1}{3} × 12 - \frac{1}{4} × 12 + \frac{1}{12} × 12$
$= 2$。
故原式$= \frac{1}{2}$。
请仿照上述方法计算：
$( -\frac{1}{42} ) ÷ ( \frac{3}{7} - \frac{5}{21} + \frac{2}{3} - \frac{1}{6} )$。

原式的倒数为 $(\dfrac{3}{7}-\dfrac{5}{21}+\dfrac{2}{3}-\dfrac{1}{6}) ÷ (-\dfrac{1}{42}) =(\dfrac{3}{7}-\dfrac{5}{21}+\dfrac{2}{3}-\dfrac{1}{6})×(-42)=\dfrac{3}{7}×(-42)-\dfrac{5}{21}×(-42)+\dfrac{2}{3}×(-42)-\dfrac{1}{6}×(-42)=-18+10-28+7=-29.$ 故原式$=-\dfrac{1}{29}.$

5. 计算：$(-\dfrac{1}{24})÷(1 - 1\dfrac{1}{2} - \dfrac{3}{8} + \dfrac{7}{12}).$

$(1-1\dfrac{1}{2}-\dfrac{3}{8}+\dfrac{7}{12})÷(-\dfrac{1}{24})=(1-\dfrac{3}{2}-\dfrac{3}{8}+\dfrac{7}{12})×(-24)=-24+36+9-14=7.$ 故原式$=\dfrac{1}{7}.$