《2025年初中必刷题七年级数学上册人教版》

8 [较难]计算:$(-\frac {1}{2}×\frac {3}{2})×(-\frac {2}{3}×\frac {4}{3})×(-\frac {3}{4}×\frac {5}{4})×... ×(-\frac {2021}{2022}×\frac {2023}{2022})$.

答案： 【解】原式=-$\frac{1}{2}$×($\frac{3}{2}$×$\frac{2}{3}$)×($\frac{4}{3}$×$\frac{3}{4}$)×($\frac{5}{4}$×$\frac{4}{5}$)×…×($\frac{2022}{2021}$×$\frac{2021}{2022}$)×$\frac{2023}{2022}$=-$\frac{1}{2}$×1×1×1×…×1×$\frac{2023}{2022}$=-$\frac{2023}{4044}$.

9 用简便方法计算:
(1) [2025山东济宁期中,中]$(-2024\frac {5}{6})+4046\frac {2}{3}+(-2025\frac {2}{3})+1\frac {5}{6}$.
(2) [2025河南郑州期中,中]$(-199\frac {37}{38})×76$.

答案： 【解】
(1)(-2024$\frac{5}{6}$)+4046$\frac{2}{3}$+(-2025$\frac{2}{3}$)+1$\frac{5}{6}$=[(-2024)+(-$\frac{5}{6}$)]+(4046+$\frac{2}{3}$)+[(-2025)+(-$\frac{2}{3}$)]+(1+$\frac{5}{6}$)=[(-2024)+4046+(-2025)+1]+[(-$\frac{5}{6}$)+$\frac{2}{3}$+(-$\frac{2}{3}$)+$\frac{5}{6}$]=-2+0=-2.
(2)(-199$\frac{37}{38}$)×76=(-200+$\frac{1}{38}$)×76=-200×76+$\frac{1}{38}$×76=-15200+2=-15198.

(1)$\frac {1}{n(n+1)}=$$\frac{1}{n}$-$\frac{1}{n+1}$(n为正整数);
(2)$1-\frac {1}{2}-\frac {1}{6}-\frac {1}{12}-\frac {1}{20}-\frac {1}{30}-\frac {1}{42}$;
【解】$1-\frac {1}{2}-\frac {1}{6}-\frac {1}{12}-\frac {1}{20}-\frac {1}{30}-\frac {1}{42}=1-\frac {1}{1×2}-\frac {1}{2×3}-\frac {1}{3×4}-\frac {1}{4×5}-\frac {1}{5×6}-\frac {1}{6×7}=1-1+\frac {1}{2}-\frac {1}{2}+\frac {1}{3}-…-\frac {1}{6}+\frac {1}{7}=\frac{1}{7}$.
(3)$\frac {1}{2×4}+\frac {1}{4×6}+\frac {1}{6×8}+... +\frac {1}{2018×2020}$.
【解】原式=$\frac{1}{4}$×(1-$\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+…+$\frac{1}{1009}$-$\frac{1}{1010}$)=$\frac{1}{4}$×(1-$\frac{1}{1010}$)=$\frac{1}{4}$×$\frac{1009}{1010}$=$\frac{1009}{4040}$.

答案：
(1)$\frac{1}{n}$-$\frac{1}{n+1}$
【解】
(2)1-$\frac{1}{2}$-$\frac{1}{6}$-$\frac{1}{12}$-$\frac{1}{20}$-$\frac{1}{30}$-$\frac{1}{42}$=1-$\frac{1}{1×2}$-$\frac{1}{2×3}$-$\frac{1}{3×4}$-$\frac{1}{4×5}$-$\frac{1}{5×6}$-$\frac{1}{6×7}$=1-1+$\frac{1}{2}$-$\frac{1}{2}$+$\frac{1}{3}$-…-$\frac{1}{6}$+$\frac{1}{7}$=$\frac{1}{7}$.
(3)原式=$\frac{1}{4}$×(1-$\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+…+$\frac{1}{1009}$-$\frac{1}{1010}$)=$\frac{1}{4}$×(1-$\frac{1}{1010}$)=$\frac{1}{4}$×$\frac{1009}{1010}$=$\frac{1009}{4040}$.

计算:$50÷(\frac {1}{3}-\frac {1}{4}+\frac {1}{12})$.
解法1:原式$=50÷\frac {1}{3}-50÷\frac {1}{4}+50÷\frac {1}{12}= 50×3-50×4+50×12$.该解法对吗?答:不对.
解法2:先计算原式的倒数,$(\frac {1}{3}-\frac {1}{4}+\frac {1}{12})÷50= \frac {1}{3}×\frac {1}{50}-\frac {1}{4}×\frac {1}{50}+\frac {1}{12}×\frac {1}{50}= \frac {1}{300}$,故原式= 300.
(1)请你用解法2的方法计算:$(-\frac {1}{30})÷(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})$;
【解】先计算原式的倒数,($\frac{2}{3}$-$\frac{1}{10}$+$\frac{1}{6}$-$\frac{2}{5}$)÷(-$\frac{1}{30}$)=$\frac{2}{3}$×(-30)-$\frac{1}{10}$×(-30)+$\frac{1}{6}$×(-30)-$\frac{2}{5}$×(-30)=-20-(-3)+(-5)-(-12)=-20+3-5+12=-10,故原式=-$\frac{1}{10}$.
(2)计算:$(1\frac {3}{4}-\frac {7}{8}-\frac {7}{12})÷(-\frac {7}{8})+(-\frac {7}{8})÷(1\frac {3}{4}-\frac {7}{8}-\frac {7}{12})$,这个题简单吗?来吧,试试吧!
【解】(1$\frac{3}{4}$-$\frac{7}{8}$-$\frac{7}{12}$)÷(-$\frac{7}{8}$)=$\frac{7}{4}$×(-$\frac{8}{7}$)-$\frac{7}{8}$×(-$\frac{8}{7}$)-$\frac{7}{12}$×(-$\frac{8}{7}$)=-2-(-1)-(-$\frac{2}{3}$)=-2+1+$\frac{2}{3}$=-$\frac{1}{3}$,所以(-$\frac{7}{8}$)÷(1$\frac{3}{4}$-$\frac{7}{8}$-$\frac{7}{12}$)=-3,所以原式=-$\frac{1}{3}$+(-3)=-$\frac{10}{3}$.

答案： 【解】因为除法没有分配律,所以解法1不对.故答案为不对.
(1)先计算原式的倒数,($\frac{2}{3}$-$\frac{1}{10}$+$\frac{1}{6}$-$\frac{2}{5}$)÷(-$\frac{1}{30}$)=$\frac{2}{3}$×(-30)-$\frac{1}{10}$×(-30)+$\frac{1}{6}$×(-30)-$\frac{2}{5}$×(-30)=-20-(-3)+(-5)-(-12)=-20+3-5+12=-10,故原式=-$\frac{1}{10}$.
(2)(1$\frac{3}{4}$-$\frac{7}{8}$-$\frac{7}{12}$)÷(-$\frac{7}{8}$)=$\frac{7}{4}$×(-$\frac{8}{7}$)-$\frac{7}{8}$×(-$\frac{8}{7}$)-$\frac{7}{12}$×(-$\frac{8}{7}$)=-2-(-1)-(-$\frac{2}{3}$)=-2+1+$\frac{2}{3}$=-$\frac{1}{3}$,所以(-$\frac{7}{8}$)÷(1$\frac{3}{4}$-$\frac{7}{8}$-$\frac{7}{12}$)=-3,所以原式=-$\frac{1}{3}$+(-3)=-$\frac{10}{3}$.