《2025年名校课堂七年级数学上册人教版陕西专版》

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

答案： 【例5】解：原式$=(1-2-3+4)+(5-6-7+8)+\cdots+(97-98-99+100)=0+0+\cdots+0=0.$

强化训练
计算（能用简便方法计算的尽量用简便方法）：
(1)$-9 + 6 - (+11) - (-15)$.
(2)$3.5 - 4.6 + 3.5 - 2.4$.
(3)$-4\frac{7}{8} - (-5\frac{1}{2}) + (-4\frac{1}{2}) - 3\frac{1}{8}$.
(4)$0 - 21\frac{2}{3} + (+3\frac{1}{4}) - (-\frac{2}{3}) - (+ \frac{1}{4})$.
(5)$|-2\frac{1}{4}| - (-\frac{3}{4}) + 1 - |1 - \frac{1}{2}|$.
(6)$(-102\frac{1}{6}) - (-96\frac{1}{2}) + 54\frac{2}{3} + (-48\frac{3}{4})$.
(7)$1 + 2 + 3 + \cdots + 2024 + (-1) + (-2) + (-3) + \cdots + (-2025)$.
(8)$\frac{1}{1×4} + \frac{1}{4×7} + \frac{1}{7×10} + \cdots + \frac{1}{301×304}$.

答案： 强化训练
解：
(1)原式=-9+6-11+15=(-9-11)+(6+15)=-20+21=1.
(2)原式=(3.5+3.5)+(-4.6-2.4)=7+(-7)=0.
(3)原式$=-4\frac{7}{8}+5\frac{1}{2}-4\frac{1}{2}$
$=-3\frac{1}{8}=(-4\frac{7}{8}-3\frac{1}{8})+(5\frac{1}{2}-4\frac{1}{2})=-8+1=-7.$
(4)原式=-21
(5)原式$=\frac{2}{3}+3\frac{1}{4}+\frac{2}{3}+\frac{1}{4}=(-21\frac{2}{3}+\frac{2}{3})+(3\frac{1}{4}+\frac{1}{4})=-21+3\frac{1}{2}=-18.$
(5)原式$=2\frac{1}{4}+3\frac{3}{4}+1-\frac{1}{2}=3\frac{1}{2}.$
(6)原式$=(-102\frac{1}{6})+96\frac{1}{2}+54\frac{2}{3}+(-48\frac{3}{4})=[(-102)+(-\frac{1}{6})]+(96+\frac{1}{2})+(54+\frac{2}{3})+[(-48)+(-\frac{3}{4})]=[(-102)+96+54+(-48)]+[(-\frac{1}{6})+\frac{1}{2}+\frac{2}{3}+(-\frac{3}{4})]=0+\frac{1}{4}=\frac{1}{4}.$
(7)原式$=(1-1)+(2-2)+(3-3)+\cdots+(2024-2024)+(-2025)=0÷0+0+\cdots+0-2025=-2025.$
(8)原式$=\frac{1}{3}×(1-\frac{1}{4})+\frac{1}{3}×(\frac{1}{4}-\frac{1}{7})+\frac{1}{3}×(\frac{1}{7}-\frac{1}{10})+\cdots+\frac{1}{3}×(\frac{1}{301}-\frac{1}{304})=\frac{1}{3}×(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\cdots+\frac{1}{301}-\frac{1}{304})=\frac{1}{3}×(1-\frac{1}{304})=\frac{101}{304}$