【典例2】计算：
(1)$(-81)÷ \dfrac{9}{4}× \dfrac{4}{9};$
(2)$32÷ 4× (-\dfrac{1}{4})÷ (-0.5).$

(1)$(-81)÷ \dfrac{9}{4}× \dfrac{4}{9}$
$=(-81)× \dfrac{4}{9}× \dfrac{4}{9}$
$=-16$；
(2)$32÷ 4× (-\dfrac{1}{4})÷ (-0.5)$
$=32× \dfrac{1}{4}× (-\dfrac{1}{4})÷ (-\dfrac{1}{2})$
$=32× \dfrac{1}{4}× \dfrac{1}{4}× 2$
$=4$.

1. 计算$1÷(-3)×(-\frac{1}{3})$的结果是 （）

A.$1$
B.$\frac{1}{9}$
C.$-1$
D.$-\frac{1}{9}$

B 原式$=1× (-\dfrac{1}{3})× (-\dfrac{1}{3})=\dfrac{1}{9}$.

2.(2025·盐城期中)计算$\frac{1}{3}×3÷\frac{1}{3}×3$的结果是。

【解析】原式$=\dfrac{1}{3}× 3× 3× 3=9$.
答案:9

3.(2025·北京期中)计算:
(1)$-2.5÷\frac{5}{32}×(-\frac{1}{8})÷(-4)$.
(2)$(\frac{1}{3}-\frac{11}{21}+\frac{3}{14})÷(-\frac{1}{42})$.

【解析】(1)原式$=-\dfrac{5}{2}× \dfrac{32}{5}× (-\dfrac{1}{8})× (-\dfrac{1}{4})$
$=-\dfrac{5}{2}× \dfrac{32}{5}× \dfrac{1}{8}× \dfrac{1}{4}$
$=-\dfrac{1}{2}$.
(2)原式$=(\dfrac{1}{3}-\dfrac{11}{21}+\dfrac{3}{14})× (-42)$
$=\dfrac{1}{3}× (-42)-\dfrac{11}{21}× (-42)+\dfrac{3}{14}× (-42)$
$=-14+22-9$
$=-1$.

【解决问题】
(1)若$f(n,-2)=(-2)÷(-2)÷(-2)÷(-2)$，则$n=$；
(2)$f(5,\dfrac{1}{3})=$；
(3)计算：$f(2,\dfrac{1}{4})÷ f(3,-3)× f(4,\dfrac{1}{2})× f(1,-1)$。

【解析】(1)因为 $f(n,-2)=(-2)÷ (-2)÷ (-2)÷ (-2)$，
所以 $n=4$；
答案:4
(2)$f(5,\dfrac{1}{3})=\dfrac{1}{3}÷ \dfrac{1}{3}÷ \dfrac{1}{3}÷ \dfrac{1}{3}÷ \dfrac{1}{3}=\dfrac{1}{3}× 3× 3× 3× 3=27$；
答案:27
(3)原式$=(\dfrac{1}{4}÷ \dfrac{1}{4})÷ [(-3)÷ (-3)÷ (-3)]× (\dfrac{1}{2}÷ \dfrac{1}{2}÷ \dfrac{1}{2}÷ \dfrac{1}{2})× (-1)$
$=1÷ (-\dfrac{1}{3})× 4× (-1)$
$=1× (-3)× 4× (-1)$
$=12$.