《2025年课时提优计划作业本七年级数学上册苏科版》

6. 计算:
(1)$(-3)-|-8|-2×(-4);$
(2)$-1^{4}-\frac {1}{2}×[3-(-3)^{2}];$
(3)$24×(-99\frac {47}{48});$
(4)$-1^{2026}-(1-0.5)^{2}×\frac {1}{5}×[2+(-3^{2})];$
(5)$1\frac {1}{2}×\frac {5}{7}-(-\frac {5}{7})×2\frac {1}{2}+(-\frac {1}{2})÷1\frac {2}{5};$
(6)$[1\frac {2}{13}-(\frac {5}{8}-\frac {1}{6}+\frac {7}{12})×24]÷(-5).$

答案： 6.
(1)原式$=-3-8-(-8)=-3$.
(2)原式$=-1-\frac {1}{2}×(3-9)=-1-\frac {1}{2}×(-6)=-1+3=2$.
(3)原式$=24×(-100+\frac {1}{48})=-24×100+24×\frac {1}{48}=-2400+\frac {1}{2}=-2399\frac {1}{2}$.
(4)原式$=-1-\frac {1}{4}×\frac {1}{5}×[2+(-27)]=-1-\frac {1}{20}×(-25)=-1+\frac {5}{4}=\frac {1}{4}$.
(5)原式$=\frac {3}{2}×\frac {5}{7}+\frac {5}{7}×\frac {5}{2}-\frac {1}{2}×\frac {5}{7}=\frac {5}{7}×(\frac {3}{2}+\frac {5}{2}-\frac {1}{2})=\frac {5}{7}×\frac {7}{2}=\frac {5}{2}$.
(6)原式$=[\frac {15}{13}-(15-4+14)]×(-\frac {1}{5})=(\frac {15}{13}-25)×(-\frac {1}{5})=-\frac {3}{13}-(-5)=\frac {62}{13}$.

7. 我们知道$a÷b= \frac {a}{b},b÷a= \frac {b}{a}$,显然$a÷b与b÷a$的结果互为倒数关系.小明利用这一思想方法计算$(-\frac {1}{30})÷(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})$的过程如下:因为$(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})÷(-\frac {1}{30})= (\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})×(-30)= -20+3-5+12= -10$,所以$(-\frac {1}{30})÷(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})= -\frac {1}{10}.$请你仿照这种方法计算:$(-\frac {1}{42})÷(\frac {1}{6}-\frac {3}{14}+\frac {2}{3}-\frac {2}{7}).$

答案： 7. 因为$(\frac {1}{6}-\frac {3}{14}+\frac {2}{3}-\frac {2}{7})÷(-\frac {1}{42})=(\frac {1}{6}-\frac {3}{14}+\frac {2}{3}-\frac {2}{7})×(-42)=-7+9-28+12=-14$,所以$(-\frac {1}{42})÷(\frac {1}{6}-\frac {3}{14}+\frac {2}{3}-\frac {2}{7})=-\frac {1}{14}.$