《2025年同步测控优化设计七年级数学上册人教版福建专版》

22. 数学老师布置了一道思考题：“计算$(-\dfrac{1}{60})÷(\dfrac{1}{3}-\dfrac{3}{10}+\dfrac{1}{12}-\dfrac{1}{5})$”,小红和小明两位同学经过仔细思考,用不同的方法解答了这个问题.
小红的解法：原式的倒数为$(\dfrac{1}{3}-\dfrac{3}{10}+\dfrac{1}{12}-\dfrac{1}{5})÷(-\dfrac{1}{60})= (\dfrac{1}{3}-\dfrac{3}{10}+\dfrac{1}{12}-\dfrac{1}{5})×(-60)= -20+18-5+12= 5$,所以$(-\dfrac{1}{60})÷(\dfrac{1}{3}-\dfrac{3}{10}+\dfrac{1}{12}-\dfrac{1}{5})= \dfrac{1}{5}$.
小明的解法：原式$=(-\dfrac{1}{60})÷[(\dfrac{1}{3}+\dfrac{1}{12})-(\dfrac{3}{10}+\dfrac{1}{5})]= (-\dfrac{1}{60})÷(\dfrac{5}{12}-\dfrac{1}{2})= -\dfrac{1}{60}×(-12)= \dfrac{1}{5}$.
请你分别用小红和小明的方法计算：$(-\dfrac{1}{24})÷(\dfrac{1}{6}-\dfrac{5}{12}+\dfrac{3}{8}-\dfrac{2}{3})$.

答案： 解法1:原式的倒数为$(\frac {1}{6}-\frac {5}{12}+\frac {3}{8}-\frac {2}{3})÷(-\frac {1}{24})=(\frac {1}{6}-\frac {5}{12}+\frac {3}{8}-\frac {2}{3})×(-24)=-4+10-9+16=13,$所以$(-\frac {1}{24})÷(\frac {1}{6}-\frac {5}{12}+\frac {3}{8}-\frac {2}{3})$的值为$\frac {1}{13}.$法2:原式$=(-\frac {1}{24})÷[(\frac {1}{6}+\frac {3}{8})-(\frac {5}{12}+\frac {2}{3})]$$=(-\frac {1}{24})÷(\frac {13}{24}-\frac {13}{12})=(-\frac {1}{24})÷(-\frac {13}{24})$$=-\frac {1}{24}×(-\frac {24}{13})=\frac {1}{13}.$

23. 小明有$5$张写着不同数字的卡片,请你按要求抽出卡片,完成下列问题：
$\boxed{(-3)}$ $\boxed{(-5)}$ $\boxed{0}$ $\boxed{+3}$ $\boxed{+4}$

(1)从中取出$2$张卡片,使这$2$张卡片上数字的乘积最小,乘积的最小值为______；
(2)从中取出$4$张卡片,用学过的运算方法,使结果为$24$. 写出运算式子.(写出一种即可)
计算结果为$24$的式子为______.

答案：
(1)-20
(2)$[-3-(-5)]×[(+3)×(+4)]=24$(答案不唯一) 解析
(1)$(+4)×(-5)=-20$,所以乘积的最小值是-20.
(2)$[-3-(-5)]×[(+3)×(+4)]=(-3+5)×12=24.$