1. 计算：$1\frac {9}{10}+2\frac {8}{10}+3\frac {7}{10}+... +8\frac {2}{10}+9\frac {1}{10}.$
答案：解:原式=(1+2+3+…+8+9)+($\frac{9}{10}+\frac{8}{10}+\frac{7}{10}+\dots +\frac{2}{10}+\frac{1}{10}$)
=45+$\frac{45}{10}$
=$49\frac{1}{2}$.
=45+$\frac{45}{10}$
=$49\frac{1}{2}$.
2. 计算：$\frac {1}{2}+\frac {5}{6}+\frac {11}{12}+\frac {19}{20}+\frac {29}{30}+... +\frac {9701}{9702}+\frac {9899}{9900}.$
答案：解:原式=$1-\frac{1}{1× 2}+1-\frac{1}{2× 3}+1-\frac{1}{3× 4}+\dots +1-\frac{1}{98× 99}+1-\frac{1}{99× 100}$
=99-$(1-\frac{1}{2})-(\frac{1}{2}-\frac{1}{3})-(\frac{1}{3}-\frac{1}{4})-\dots -(\frac{1}{98}-\frac{1}{99})-(\frac{1}{99}-\frac{1}{100})$
=99-$(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\dots +\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100})$
=99-$(1-\frac{1}{100})$
=99-1+$\frac{1}{100}$
=$98\frac{1}{100}$.
=99-$(1-\frac{1}{2})-(\frac{1}{2}-\frac{1}{3})-(\frac{1}{3}-\frac{1}{4})-\dots -(\frac{1}{98}-\frac{1}{99})-(\frac{1}{99}-\frac{1}{100})$
=99-$(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\dots +\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100})$
=99-$(1-\frac{1}{100})$
=99-1+$\frac{1}{100}$
=$98\frac{1}{100}$.
3. (2024秋·拱墅区月考)阅读下面的解题过程,并用解题过程中的解题方法解决问题.
示例:计算:$(-1\frac {5}{6})+(-2\frac {2}{3})+9\frac {3}{4}+(-3\frac {1}{2}).$
解:原式$=[(-1)+(-\frac {5}{6})]+[(-2)+(-\frac {2}{3})]+(9+\frac {3}{4})+[(-3)+(-\frac {1}{2})]$
$=[(-1)+(-2)+9+(-3)]+[(-\frac {5}{6})+(-\frac {2}{3})+\frac {3}{4}+(-\frac {1}{2})]= 3+(-\frac {5}{4})= \frac {7}{4}.$
以上解题方法叫拆项法.
请你利用拆项法计算下面式子的值.
$(-2024\frac {5}{6})+(-2022\frac {2}{3})+(-1\frac {1}{2})+(-\frac {5}{6})+4046\frac {3}{4}.$
示例:计算:$(-1\frac {5}{6})+(-2\frac {2}{3})+9\frac {3}{4}+(-3\frac {1}{2}).$
解:原式$=[(-1)+(-\frac {5}{6})]+[(-2)+(-\frac {2}{3})]+(9+\frac {3}{4})+[(-3)+(-\frac {1}{2})]$
$=[(-1)+(-2)+9+(-3)]+[(-\frac {5}{6})+(-\frac {2}{3})+\frac {3}{4}+(-\frac {1}{2})]= 3+(-\frac {5}{4})= \frac {7}{4}.$
以上解题方法叫拆项法.
请你利用拆项法计算下面式子的值.
$(-2024\frac {5}{6})+(-2022\frac {2}{3})+(-1\frac {1}{2})+(-\frac {5}{6})+4046\frac {3}{4}.$
答案：解:$(-2024\frac{5}{6})+(-2022\frac{2}{3})+(-1\frac{1}{2})+(-\frac{5}{6})+4046\frac{3}{4}$
=$[(-2024)+(-\frac{5}{6})]+[(-2022)+(-\frac{2}{3})]+[(-1)+(-\frac{1}{2})]+(-\frac{5}{6})+(4046+\frac{3}{4})$
=$[(-2024)+(-2022)+(-1)+4046]+[(-\frac{5}{6})+(-\frac{2}{3})+(-\frac{1}{2})+(-\frac{5}{6})+\frac{3}{4}]$
=-1+$(-\frac{25}{12})$
=$-\frac{37}{12}$.
=$[(-2024)+(-\frac{5}{6})]+[(-2022)+(-\frac{2}{3})]+[(-1)+(-\frac{1}{2})]+(-\frac{5}{6})+(4046+\frac{3}{4})$
=$[(-2024)+(-2022)+(-1)+4046]+[(-\frac{5}{6})+(-\frac{2}{3})+(-\frac{1}{2})+(-\frac{5}{6})+\frac{3}{4}]$
=-1+$(-\frac{25}{12})$
=$-\frac{37}{12}$.