《2025年实验班提优训练七年级数学上册人教版》

12. (2025·广东深圳期中)a,b,c为有理数.
(1)如果$ab＞0,a+b＞0$,试确定a,b的正负；
(2)如果$ab＞0,abc＞0,bc＜0$,试确定a,b,c的正负.

答案：
(1)$\because ab\gt0$,$\therefore a$,$b$同号.$\because a+b\gt0$,$\therefore a$,$b$都为正数.
(2)$\because ab\gt0$,$\therefore a$,$b$同号.$\because abc\gt0$,$\therefore c\gt0$.又$bc\lt0$,$\therefore b$,$c$异号,即$b\lt0$.故$a\lt0$.$\therefore a$,$b$为负数,$c$为正数.归纳总结 本题综合考查了有理数的加法和乘法运算,解答本题的关键是熟悉并灵活应用有理数的乘法法则.

13. 整体思想 中考新考法 解题方法型阅读理解题 (2025·安徽淮北期末)阅读理解：
计算$(1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4})×(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5})-(1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5})×(\frac {1}{2}+\frac {1}{3}+\frac {1}{4})$时,若把$(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5})与(\frac {1}{2}+\frac {1}{3}+\frac {1}{4})$分别看作一个整体,再利用分配律进行运算,可以降低难度. 过程如下：
解：设$(\frac {1}{2}+\frac {1}{3}+\frac {1}{4})$为A,$(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5})$为B,则原式$=B(1+A)-A(1+B)= B+AB-A-AB= B-A= \frac {1}{5}$. 请用上面方法计算：
(1)$(1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6})(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7})-(1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7})\cdot (\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6})$；
(2)$(1+\frac {1}{2}+\frac {1}{3}+... +\frac {1}{n})(\frac {1}{2}+\frac {1}{3}+... +\frac {1}{n+1})-(1+\frac {1}{2}+\frac {1}{3}+... +\frac {1}{n+1})(\frac {1}{2}+\frac {1}{3}+... +\frac {1}{n})$.

答案：
(1)设$\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)$为$A$,$\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)$为$B$,原式$=(1+A)B-(1+B)A=B+AB-A-AB=B-A=\frac{1}{7}$.
(2)设$\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots+\frac{1}{n}\right)$为$A$,$\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots+\frac{1}{n+1}\right)$为$B$,原式$=(1+A)B-(1+B)A=B+AB-A-AB=B-A=\frac{1}{n+1}$.归纳总结 此题考查了有理数的乘法,熟练掌握阅读理解中的解题方法是解题的关键.