《2025年经纶学典5星学霸七年级数学上册苏科版》

1. 计算:
(1)$(-2.01)+2.5+1+2.01+(-\frac {5}{2})$;
(2)$(-\frac {3}{7})×(-\frac {4}{5})×(-\frac {7}{12})$.

答案： 1.
(1)原式=(-2.01)+2.01+2.5+(-$\frac{5}{2}$)+1=1.
(2)原式=(-$\frac{3}{7}$)×(-$\frac{7}{12}$)×(-$\frac{4}{5}$)=$\frac{1}{4}$×(-$\frac{4}{5}$)=-$\frac{1}{5}$.

2. 计算$(\frac {11}{12}-\frac {7}{6}+\frac {3}{4}-\frac {13}{24})×(-24)$的结果是 (A)
A.1
B.-1
C.10
D.-10

答案： 2. A

3. 计算:(1)$17.48×37-174.8×1.9+1.748×820=$1748;
(2)$999×118\frac {4}{5}+333×(-\frac {3}{5})-999×18\frac {3}{5}=$99900.

答案： 3.
(1)1 748
(2)99 900

4. (2025·连云港校级月考)阅读下列材料:
计算:$\frac {1}{60}÷(\frac {1}{3}-\frac {1}{4}+\frac {1}{12})$.
解法①:原式$=\frac {1}{60}÷\frac {1}{3}-\frac {1}{60}÷\frac {1}{4}+\frac {1}{60}÷\frac {1}{12}= \frac {1}{60}×3-\frac {1}{60}×4+\frac {1}{60}×12= \frac {11}{60}$.
解法②:原式的倒数为$(\frac {1}{3}-\frac {1}{4}+\frac {1}{12})÷\frac {1}{60}= (\frac {1}{3}-\frac {1}{4}+\frac {1}{12})×60= \frac {1}{3}×60-\frac {1}{4}×60+\frac {1}{12}×60= 20-15+5= 10$,
所以原式$=\frac {1}{10}$.
(1)上述解法中,你认为解法____①____是错误的;
(2)计算:$-\frac {1}{42}÷(\frac {3}{7}-\frac {5}{14}+\frac {2}{3}-\frac {1}{6})$.
原式的倒数为($\frac{3}{7}$-$\frac{5}{14}$+$\frac{2}{3}$-$\frac{1}{6}$)÷(-$\frac{1}{42}$)=($\frac{3}{7}$-$\frac{5}{14}$+$\frac{2}{3}$-$\frac{1}{6}$)×(-42)=$\frac{3}{7}$×(-42)-$\frac{5}{14}$×(-42)+$\frac{2}{3}$×(-42)-$\frac{1}{6}$×(-42)=-18+15-28+7=-24,所以原式=-$\frac{1}{24}$.

答案： 4.
(1)① 解析:解法①是错误的,除法没有分配律.
(2)原式的倒数为($\frac{3}{7}$-$\frac{5}{14}$+$\frac{2}{3}$-$\frac{1}{6}$)÷(-$\frac{1}{42}$)=($\frac{3}{7}$-$\frac{5}{14}$+$\frac{2}{3}$-$\frac{1}{6}$)×(-42)=$\frac{3}{7}$×(-42)-$\frac{5}{14}$×(-42)+$\frac{2}{3}$×(-42)-$\frac{1}{6}$×(-42)=-18+15-28+7=-24,所以原式=-$\frac{1}{24}$.