3. $1 - 0.5 - \frac{1}{6} - \frac{1}{12} - 0.05 - \frac{1}{30}$ 的结果是多少？

$1 - 0.5 - \frac{1}{6} - \frac{1}{12} - 0.05 - \frac{1}{30}=1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}=1-(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30})=1-(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\dots+\frac{1}{5}-\frac{1}{6})=1-(1-\frac{1}{6})=\frac{1}{6}$

例② 用简便方法计算$\frac{3}{4}+9\frac{3}{4}+99\frac{3}{4}+999\frac{3}{4}$。
思路点拨
观察式子发现$\frac{3}{4}=1-\frac{1}{4},9\frac{3}{4}=10-\frac{1}{4},99\frac{3}{4}=100-\frac{1}{4},999\frac{3}{4}=1000-\frac{1}{4}$,根据加法的交换律、结合律以及减法的性质进行简便计算。
规范解答

$\frac{3}{4}+9\frac{3}{4}+99\frac{3}{4}+999\frac{3}{4}$
$=1-\frac{1}{4}+10-\frac{1}{4}+100-\frac{1}{4}+1000-\frac{1}{4}$
$=(1+10+100+1000)-(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4})$
$=1111-1$
$=1110$

$\frac{3}{4}+9\frac{3}{4}+99\frac{3}{4}+999\frac{3}{4}$
$=1-\frac{1}{4}+10-\frac{1}{4}+100-\frac{1}{4}+1000-\frac{1}{4}$
$=(1+10+100+1000)-(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4})$
$=1111-1$
$=1110$

1. 用简便方法计算下面各题。
（1）$11\frac{1}{5} + 101\frac{1}{5} + 1001\frac{1}{5} + 10001\frac{1}{5}$
（2）$\frac{49}{5} + \frac{499}{5} + \frac{4999}{5} + \frac{3}{5}$

（1）$11\frac{1}{5}+101\frac{1}{5}+1001\frac{1}{5}+10001\frac{1}{5}=10+100+1000+10000+1\frac{1}{5}+1\frac{1}{5}+1\frac{1}{5}+1\frac{1}{5}=11110+4\frac{4}{5}=11114\frac{4}{5}$
（2）$\frac{49}{5}+\frac{499}{5}+\frac{4999}{5}+\frac{3}{5}=10-\frac{1}{5}+100-\frac{1}{5}+1000-\frac{1}{5}+\frac{3}{5}=1110$

2. $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}+\frac{127}{128}$的整数部分是多少？

$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}+\frac{127}{128}=1-\frac{1}{2}+1-\frac{1}{4}+1-\frac{1}{8}+1-\frac{1}{16}+1-\frac{1}{32}+1-\frac{1}{64}+1-\frac{1}{128}=7-(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128})=7-\frac{127}{128}=6\frac{1}{128}$
答:整数部分是6。