《2025年名校课堂七年级数学上册北师大版安徽专版》

【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) $ 1.4 - (-3.6) - 5.2 - 4.3 - (-1.5) $.
(5) $ (-102\dfrac{1}{6}) - (-96\dfrac{1}{2}) + 54\dfrac{2}{3} + (-48\dfrac{3}{4}) $.
(6) $ 1 + 2 + 3 + \cdots + 2024 + (-1) + (-2) + (-3) + \cdots + (-2025) $.
(7) $ \vert \dfrac{1}{2} - 1 \vert + \vert \dfrac{1}{3} - \dfrac{1}{2} \vert + \vert \dfrac{1}{4} - \dfrac{1}{3} \vert + \cdots + \vert \dfrac{1}{2025} - \dfrac{1}{2024} \vert $.
(8) $ \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=1.4+3.6-5.2-4.3+1.5=(1.4+3.6)+(-5.2-4.3+1.5)=5-8=-3.
(5)原式$=(-102\frac {1}{6})+96\frac {1}{2}+54\frac {2}{3}+(-48\frac {3}{4})=[(-102)+(-\frac {1}{6})]+(96+\frac {1}{2})+(54+\frac {2}{3})+[(-48)+(-\frac {3}{4})]=[(-102)+96+54+(-48)]+[(-\frac {1}{6})+\frac {1}{2}+\frac {2}{3}+(-\frac {3}{4})]=0+\frac {1}{4}=\frac {1}{4}.(6)$原式=(1-1)+(2-2)+(3-3)+…+(2024-2024)+(-2025)=0+0+0+…+0-2025=-2025.
(7)原式$=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+…+\frac {1}{2024}-\frac {1}{2025}=1-\frac {1}{2025}=\frac {2024}{2025}.(8)$原式$=\frac {1}{3}×(1-\frac {1}{4})+\frac {1}{3}×(\frac {1}{4}-\frac {1}{7})+\frac {1}{3}×(\frac {1}{7}-\frac {1}{10})+…+\frac {1}{3}×(\frac {1}{301}-\frac {1}{304})=\frac {1}{3}×[(1-\frac {1}{4})+(\frac {1}{4}-\frac {1}{7})+(\frac {1}{7}-\frac {1}{10})+…+(\frac {1}{301}-\frac {1}{304})]=\frac {1}{3}×(1-\frac {1}{304})=\frac {101}{304}.$