《2025年全品学练考七年级数学上册华师大版》

7. 计算：$-28×(-0.125) + (-28)×\frac{1}{8} - 28×(-\frac{4}{7})$.

答案： 7. 16

8. 用简便方法计算下列各题：
(1)$999×(-15)$；
(2)$999×118\frac{4}{5} + 999×(-\frac{1}{5}) - 999×18\frac{3}{5}$.

答案： 8. 解：$(1)999 × (-15) = (1000 - 1) × (-15) = -15000 + 15 = -14985. (2)999 × 18\frac{4}{5} + 999 × \left( - \frac{1}{5} \right) - 999 × 18\frac{3}{5} = 999 × \left[ 118\frac{4}{5} + \left( - \frac{1}{5} \right) - 18\frac{3}{5} \right] = 999 × 100 = 99900.$

9. 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$(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})$分别各看作一个整体，再利用分配律进行运算，可以大大简化难度. 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)$(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6})×(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}) - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7})×(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6})$；
(2)$(1 + \frac{1}{2} + \frac{1}{3} + ·s + \frac{1}{n})(\frac{1}{2} + \frac{1}{3} + ·s + \frac{1}{n + 1}) - (1 + \frac{1}{2} + \frac{1}{3} + ·s + \frac{1}{n + 1})(\frac{1}{2} + \frac{1}{3} + ·s + \frac{1}{n})(n \geq 2$且$n$为整数).

答案： 9. 解：
(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} + ·s + \frac{1}{n} \right)$为$ A, \left( \frac{1}{2} + \frac{1}{3} + ·s + \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}.$