5.计算1+2-3-4+5+6-7-8+…+2021+2022-2023-2024的结果为（）

A.0
B.-1
C.2024
D.-2024

D

6. 计算$\frac{1}{2}-(\frac{1}{2}-\frac{1}{4})-(\frac{1}{4}-\frac{1}{8})-\dots-(\frac{1}{512}-\frac{1}{1024})$的结果是。

$\dfrac{1}{1024}$

7.(新考法)阅读:因为一个非负数的绝对值等于它本身,负数的绝对值等于它的相反数,所以,当$a≥0$时,$|a|=a$;当$a＜0$时,$|a|=-a$.根据以上阅读完成:
(1)$|3.14-π|=$;
(2)计算:$|\frac{1}{2}-1|+|\frac{1}{3}-\frac{1}{2}|+|\frac{1}{4}-\frac{1}{3}|+\dots+|\frac{1}{9}-\frac{1}{8}|+|\frac{1}{10}-\frac{1}{9}|$.

(1) $π-3.14$
(2) 解:原式=$1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}$.

8.(砀山县期末)探索发现:$\frac{1}{1×2}=1-\frac{1}{2},\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3},\frac{1}{3×4}=\frac{1}{3}-\frac{1}{4},···.$
根据你发现的规律,回答下列问题:
(1)$\frac{1}{4×5}=$，
$\frac{1}{n×(n+1)}=$；
(2)计算:$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+···+\frac{1}{602×603}.$

(1) $\dfrac{1}{4}-\dfrac{1}{5}$；$\dfrac{1}{n}-\dfrac{1}{n+1}$
(2) 解: $\dfrac{1}{1×2}+\dfrac{1}{2×3}+\dfrac{1}{3×4}+…+\dfrac{1}{602×603}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{602}-\dfrac{1}{603}=1-\dfrac{1}{603}=\dfrac{602}{603}$.

9. 阅读下面的计算过程,体会“拆项法”：
计算：$-5\frac{5}{6}+(-9\frac{2}{3})+17\frac{3}{4}+(-3\frac{1}{2})$。
解：原式$=[(-5)+(-\frac{5}{6})]+[(-9)+(-\frac{2}{3})]+(17+\frac{3}{4})+[(-3)+(-\frac{1}{2})]$
$=[(-5)+(-9)+17+(-3)]+[(-\frac{5}{6})+(-\frac{2}{3})+\frac{3}{4}+(-\frac{1}{2})]$
$=0+(-1\frac{1}{4})$
$=-1\frac{1}{4}$。
启发应用：
用上面的方法完成下列计算：
(1) $(-3\frac{3}{10})+(-1\frac{1}{2})+2\frac{3}{5}-(-2\frac{1}{2})$；
(2) $(-2026\frac{5}{6})+(-2025\frac{2}{3})+4052\frac{2}{3}+(-1\frac{1}{2})$。

(1) 解:原式=$(-3-\dfrac{3}{10})+(-1-\dfrac{1}{2})+(2+\dfrac{3}{5})+(2+\dfrac{1}{2})=(-3-1+2+2)+(-\dfrac{3}{10}-\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{2})=0+\dfrac{3}{10}=\dfrac{3}{10}$.
(2) 原式=$(-2026-\dfrac{5}{6})+(-2025-\dfrac{2}{3})+(4052+\dfrac{2}{3})+(-1-\dfrac{1}{2})=(-2026-2025+4052-1)+(-\dfrac{5}{6}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{1}{2})=0-1\dfrac{1}{3}=-1\dfrac{1}{3}$.