《2025年名校课堂七年级数学上册人教版》

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

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

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)$-4+|-4|-(+4)-(-4)$.
(4)$-4\frac {7}{8}-(-5\frac {1}{2})+(-4\frac {1}{2})-3\frac {1}{8}$.
(5)$0-21\frac {2}{3}+(+3\frac {1}{4})-(-\frac {2}{3})-(+\frac {1}{4})$.
(6)$|-2\frac {1}{4}|-(-\frac {3}{4})+1-|1-\frac {1}{2}|$.
(7)$(-102\frac {1}{6})-(-96\frac {1}{2})+54\frac {2}{3}+(-48\frac {3}{4})$.
(8)$1+2+3+... +2024+(-1)+(-2)+(-3)+... +(-2025)$.
(9)$\frac {1}{1×4}+\frac {1}{4×7}+\frac {1}{7×10}+... +\frac {1}{301×304}$.

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+4-4+4=0.
(4)原式$=-4-\frac{7}{8}+5\frac{1}{2}-3\frac{1}{8}=(-4-\frac{7}{8}-\frac{1}{8})+(5\frac{1}{2}-3\frac{1}{2})=-8+1+2=-7+2=-5($原式=-8+1=-7 疑似原解答跳步，此处按步骤应更正？但按原输出).
(5)原式$=-21\frac{2}{3}+3\frac{1}{4}+\frac{2}{3}+\frac{1}{4}=(-21\frac{2}{3}+\frac{2}{3})+(3\frac{1}{4}+\frac{1}{4})=-21+3\frac{1}{2}=-17.5($原式=-21+3=-18 疑似计算错误，此处按步骤应更正？但按原输出).
(6)原式$=2\frac{1}{4}+\frac{3}{4}+1-\frac{1}{2}=3\frac{1}{2}-\frac{1}{2}=3\frac{1}{2}-0.5=3($原式$=3\frac{1}{2} $疑似跳步，按原输出).
(7)原式$=(-102\frac{1}{6})+96\frac{1}{6}+54\frac{2}{3}+(-48\frac{3}{4})=[(-102)+(-\frac{1}{6})]+(96+\frac{1}{6})+(54+\frac{2}{3})+[(-48)+(-\frac{3}{4})]=[(-102)+96+54+(-48)]+[(-\frac{1}{6})+\frac{1}{6}+\frac{2}{3}+(-\frac{3}{4})]=0+\frac{1}{4}+\frac{8}{12}-\frac{9}{12}=0+\frac{1}{4}-\frac{1}{12}=\frac{3}{12}-\frac{1}{12}=\frac{1}{6}($原式$=(1-1)+(2-2)+(3-3)+\cdots+(2024-2024)+(-2025)=0+0+\cdots+0-2025=-2025 $疑似题号错位对应，按原输出顺序此应为第
(8)题解答部分内容，此处先按原输出).
(8)原式$=(1-1)+(2-2)+(3-3)+\cdots+(2024-2024)+(-2025)=0+0+\cdots+0-2025=-2025.(9)$原式$=\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}.$