《2025年名校课堂七年级数学上册人教版福建专版》

【例 5】 计算：$1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + \cdots + 97 - 98 - 99 + 100$.

答案： 原式=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)=0+0+...+0=0.

计算（能用简便方法计算的尽量用简便方法）：
(1)$-9 + 6 - (+11) - (-15)$.
(2)$\vert -\dfrac{1}{2}\vert - (-2.5) - (-1) - \vert 0 - \dfrac{5}{2}\vert$.
(3)$\dfrac{1}{2} + (-\dfrac{2}{3}) + \dfrac{4}{5} - \dfrac{1}{2} + (-\dfrac{1}{3})$.
(4)$(-102\dfrac{1}{6}) - (-96\dfrac{1}{2}) + 54\dfrac{2}{3} + (-48\dfrac{3}{4})$.
(5)$1 + 2 + 3 + \cdots + 2024 + (-1) + (-2) + (-3) + \cdots + (-2025)$.
(6)$\dfrac{1}{1× 4} + \dfrac{1}{4× 7} + \dfrac{1}{7× 10} + \cdots + \dfrac{1}{301× 304}$.

答案：
(1)原式=-9+6-11+15=(-9-11)+(6+15)=-20+21=1.
(2)原式=$\frac {1}{2}+2.5+1-\frac {5}{2}=(2.5-\frac {5}{2})+(\frac {1}{2}+1)=\frac {3}{2}$.
(3)原式=$[\frac {1}{2}+(-\frac {1}{2})]+[(-\frac {2}{3})+(-\frac {1}{3})]+\frac {4}{5}=0+(-1)+\frac {4}{5}=-\frac {1}{5}$.
(4)原式=$(-102\frac {1}{6})+96\frac {1}{2}+54\frac {2}{3}+(-48\frac {3}{4})=[(-102)+(-\frac {1}{6})]+(96+\frac {1}{2})+(54+\frac {2}{3})+[(-48)+(-\frac {3}{4})]=[(-102)+96+54+(-48)]+[(-\frac {1}{6})+\frac {1}{2}+\frac {2}{3}+(-\frac {3}{4})]=0+\frac {1}{4}=\frac {1}{4}$.
(5)原式=(1-1)+(2-2)+(3-3)+...+(2024-2024)+(-2025)=0+0+0+...+0-2025=-2025.
(6)原式=$\frac {1}{3}×(\frac {1}{1}-\frac {1}{4})+\frac {1}{3}×(\frac {1}{4}-\frac {1}{7})+\frac {1}{3}×(\frac {1}{7}-\frac {1}{10})+... +\frac {1}{3}×(\frac {1}{301}-\frac {1}{304})=\frac {1}{3}×[(\frac {1}{1}-\frac {1}{4})+(\frac {1}{4}-\frac {1}{7})+(\frac {1}{7}-\frac {1}{10})+... +(\frac {1}{301}-\frac {1}{304})]=\frac {1}{3}×(1-\frac {1}{304})=\frac {101}{304}$.