6. $\dfrac{9}{10}+\dfrac{99}{100}+\dfrac{999}{1000}+···+\dfrac{999999}{1000000}$，这个算式结果的整数部分是(
A.4
B.5
C.6
D.7
B
)。A.4
B.5
C.6
D.7
答案：6. B
解析：
$\dfrac{9}{10}=1-\dfrac{1}{10}$，$\dfrac{99}{100}=1-\dfrac{1}{100}$，$\dfrac{999}{1000}=1-\dfrac{1}{1000}$，$···$，$\dfrac{999999}{1000000}=1-\dfrac{1}{1000000}$
原式$=(1-\dfrac{1}{10})+(1-\dfrac{1}{100})+(1-\dfrac{1}{1000})+···+(1-\dfrac{1}{1000000})$
共有$6$项，所以$=6 - (\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+···+\dfrac{1}{1000000})$
$\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+···+\dfrac{1}{1000000}=0.1+0.01+0.001+0.0001+0.00001+0.000001=0.111111$
则原式$=6 - 0.111111=5.888889$，整数部分是$5$
B
原式$=(1-\dfrac{1}{10})+(1-\dfrac{1}{100})+(1-\dfrac{1}{1000})+···+(1-\dfrac{1}{1000000})$
共有$6$项，所以$=6 - (\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+···+\dfrac{1}{1000000})$
$\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+···+\dfrac{1}{1000000}=0.1+0.01+0.001+0.0001+0.00001+0.000001=0.111111$
则原式$=6 - 0.111111=5.888889$，整数部分是$5$
B
三、计算题
1. $\dfrac{7}{12}+\dfrac{5}{12}=$ $\dfrac{3}{5}+\dfrac{9}{10}=$ $\dfrac{2}{5}-\dfrac{1}{8}-\dfrac{7}{8}+\dfrac{3}{5}=$
$\dfrac{17}{18}-\dfrac{1}{4}=$ $\dfrac{6}{7}-\dfrac{5}{6}=$ $\dfrac{2}{3}-\dfrac{2}{7}+\dfrac{1}{3}=$
1. $\dfrac{7}{12}+\dfrac{5}{12}=$ $\dfrac{3}{5}+\dfrac{9}{10}=$ $\dfrac{2}{5}-\dfrac{1}{8}-\dfrac{7}{8}+\dfrac{3}{5}=$
$\dfrac{17}{18}-\dfrac{1}{4}=$ $\dfrac{6}{7}-\dfrac{5}{6}=$ $\dfrac{2}{3}-\dfrac{2}{7}+\dfrac{1}{3}=$
答案：三、1. 1 $\dfrac{3}{2}$ 0 $\dfrac{25}{36}$ $\dfrac{1}{42}$ $\dfrac{5}{7}$
解析：
$\dfrac{7}{12}+\dfrac{5}{12}=1$；$\dfrac{3}{5}+\dfrac{9}{10}=\dfrac{6}{10}+\dfrac{9}{10}=\dfrac{15}{10}=\dfrac{3}{2}$；$\dfrac{2}{5}-\dfrac{1}{8}-\dfrac{7}{8}+\dfrac{3}{5}=(\dfrac{2}{5}+\dfrac{3}{5})-(\dfrac{1}{8}+\dfrac{7}{8})=1 - 1=0$；$\dfrac{17}{18}-\dfrac{1}{4}=\dfrac{34}{36}-\dfrac{9}{36}=\dfrac{25}{36}$；$\dfrac{6}{7}-\dfrac{5}{6}=\dfrac{36}{42}-\dfrac{35}{42}=\dfrac{1}{42}$；$\dfrac{2}{3}-\dfrac{2}{7}+\dfrac{1}{3}=(\dfrac{2}{3}+\dfrac{1}{3})-\dfrac{2}{7}=1 - \dfrac{2}{7}=\dfrac{5}{7}$
2. 计算下列各题，能简算的要简算。
$2-\dfrac{3}{14}+\dfrac{9}{14}$ $\dfrac{3}{5}-(\dfrac{1}{2}-\dfrac{1}{4})$
$\dfrac{19}{20}-(\dfrac{2}{7}+\dfrac{9}{20})$ $\dfrac{1}{6}+\dfrac{7}{8}+\dfrac{5}{6}-\dfrac{1}{8}$
$\dfrac{5}{7}-\dfrac{11}{13}+\dfrac{2}{7}-\dfrac{2}{13}$ $999\dfrac{8}{9}+99\dfrac{8}{9}+9\dfrac{8}{9}+\dfrac{1}{3}$
$4\dfrac{1}{10}+4\dfrac{11}{100}+4\dfrac{111}{1000}+4\dfrac{1111}{10000}+4\dfrac{11111}{100000}$
$2-\dfrac{3}{14}+\dfrac{9}{14}$ $\dfrac{3}{5}-(\dfrac{1}{2}-\dfrac{1}{4})$
$\dfrac{19}{20}-(\dfrac{2}{7}+\dfrac{9}{20})$ $\dfrac{1}{6}+\dfrac{7}{8}+\dfrac{5}{6}-\dfrac{1}{8}$
$\dfrac{5}{7}-\dfrac{11}{13}+\dfrac{2}{7}-\dfrac{2}{13}$ $999\dfrac{8}{9}+99\dfrac{8}{9}+9\dfrac{8}{9}+\dfrac{1}{3}$
$4\dfrac{1}{10}+4\dfrac{11}{100}+4\dfrac{111}{1000}+4\dfrac{1111}{10000}+4\dfrac{11111}{100000}$
答案：2. $2\dfrac{3}{7}$ $\dfrac{7}{20}$ $\dfrac{3}{14}$ $1\dfrac{3}{4}$ 0 1110 20.54321
解析：
$2-\dfrac{3}{14}+\dfrac{9}{14}=2+(\dfrac{9}{14}-\dfrac{3}{14})=2+\dfrac{6}{14}=2+\dfrac{3}{7}=2\dfrac{3}{7}$
$\dfrac{3}{5}-(\dfrac{1}{2}-\dfrac{1}{4})=\dfrac{3}{5}-(\dfrac{2}{4}-\dfrac{1}{4})=\dfrac{3}{5}-\dfrac{1}{4}=\dfrac{12}{20}-\dfrac{5}{20}=\dfrac{7}{20}$
$\dfrac{19}{20}-(\dfrac{2}{7}+\dfrac{9}{20})=\dfrac{19}{20}-\dfrac{9}{20}-\dfrac{2}{7}=\dfrac{10}{20}-\dfrac{2}{7}=\dfrac{1}{2}-\dfrac{2}{7}=\dfrac{7}{14}-\dfrac{4}{14}=\dfrac{3}{14}$
$\dfrac{1}{6}+\dfrac{7}{8}+\dfrac{5}{6}-\dfrac{1}{8}=(\dfrac{1}{6}+\dfrac{5}{6})+(\dfrac{7}{8}-\dfrac{1}{8})=1+\dfrac{6}{8}=1+\dfrac{3}{4}=1\dfrac{3}{4}$
$\dfrac{5}{7}-\dfrac{11}{13}+\dfrac{2}{7}-\dfrac{2}{13}=(\dfrac{5}{7}+\dfrac{2}{7})-(\dfrac{11}{13}+\dfrac{2}{13})=1-1=0$
$999\dfrac{8}{9}+99\dfrac{8}{9}+9\dfrac{8}{9}+\dfrac{1}{3}=999\dfrac{8}{9}+99\dfrac{8}{9}+9\dfrac{8}{9}+\dfrac{3}{9}=(999\dfrac{8}{9}+\dfrac{1}{9})+(99\dfrac{8}{9}+\dfrac{1}{9})+(9\dfrac{8}{9}+\dfrac{1}{9})=1000+100+10=1110$
$4\dfrac{1}{10}+4\dfrac{11}{100}+4\dfrac{111}{1000}+4\dfrac{1111}{10000}+4\dfrac{11111}{100000}$
$=4+0.1+4+0.11+4+0.111+4+0.1111+4+0.11111$
$=(4+4+4+4+4)+(0.1+0.11+0.111+0.1111+0.11111)$
$=20+0.54321=20.54321$
$\dfrac{3}{5}-(\dfrac{1}{2}-\dfrac{1}{4})=\dfrac{3}{5}-(\dfrac{2}{4}-\dfrac{1}{4})=\dfrac{3}{5}-\dfrac{1}{4}=\dfrac{12}{20}-\dfrac{5}{20}=\dfrac{7}{20}$
$\dfrac{19}{20}-(\dfrac{2}{7}+\dfrac{9}{20})=\dfrac{19}{20}-\dfrac{9}{20}-\dfrac{2}{7}=\dfrac{10}{20}-\dfrac{2}{7}=\dfrac{1}{2}-\dfrac{2}{7}=\dfrac{7}{14}-\dfrac{4}{14}=\dfrac{3}{14}$
$\dfrac{1}{6}+\dfrac{7}{8}+\dfrac{5}{6}-\dfrac{1}{8}=(\dfrac{1}{6}+\dfrac{5}{6})+(\dfrac{7}{8}-\dfrac{1}{8})=1+\dfrac{6}{8}=1+\dfrac{3}{4}=1\dfrac{3}{4}$
$\dfrac{5}{7}-\dfrac{11}{13}+\dfrac{2}{7}-\dfrac{2}{13}=(\dfrac{5}{7}+\dfrac{2}{7})-(\dfrac{11}{13}+\dfrac{2}{13})=1-1=0$
$999\dfrac{8}{9}+99\dfrac{8}{9}+9\dfrac{8}{9}+\dfrac{1}{3}=999\dfrac{8}{9}+99\dfrac{8}{9}+9\dfrac{8}{9}+\dfrac{3}{9}=(999\dfrac{8}{9}+\dfrac{1}{9})+(99\dfrac{8}{9}+\dfrac{1}{9})+(9\dfrac{8}{9}+\dfrac{1}{9})=1000+100+10=1110$
$4\dfrac{1}{10}+4\dfrac{11}{100}+4\dfrac{111}{1000}+4\dfrac{1111}{10000}+4\dfrac{11111}{100000}$
$=4+0.1+4+0.11+4+0.111+4+0.1111+4+0.11111$
$=(4+4+4+4+4)+(0.1+0.11+0.111+0.1111+0.11111)$
$=20+0.54321=20.54321$
四、解决问题
1. 参观扬州中国大运河博物馆需要预约，国庆节当天预约通道开启后的2小时，手机就显示当前预约了总票数的$\dfrac{3}{5}$，后2小时又预约了总票数的$\dfrac{8}{25}$，还剩下总票数的几分之几？
1. 参观扬州中国大运河博物馆需要预约，国庆节当天预约通道开启后的2小时，手机就显示当前预约了总票数的$\dfrac{3}{5}$，后2小时又预约了总票数的$\dfrac{8}{25}$，还剩下总票数的几分之几？
答案：四、1. $1-\dfrac{3}{5}-\dfrac{8}{25}=\dfrac{2}{25}$
2. 一块长方形桌布，长是$\dfrac{12}{5}$米，比宽多$\dfrac{1}{3}$米，这块桌布的周长是多少米？
答案：2. $\dfrac{12}{5}-\dfrac{1}{3}=\dfrac{31}{15}$（米）
$\dfrac{12}{5}+\dfrac{12}{5}+\dfrac{31}{15}+\dfrac{31}{15}=\dfrac{134}{15}$（米）
$\dfrac{12}{5}+\dfrac{12}{5}+\dfrac{31}{15}+\dfrac{31}{15}=\dfrac{134}{15}$（米）
3. 一辆货车和一辆轿车同时从A、B两地出发，相向而行。当货车行了全程的$\dfrac{4}{7}$时，轿车行了全程的$\dfrac{5}{9}$。这时两车的距离占全程的几分之几？
答案：3. $\dfrac{4}{7}+\dfrac{5}{9}-1=\dfrac{8}{63}$
解析：
$\dfrac{4}{7}+\dfrac{5}{9}-1=\dfrac{36}{63}+\dfrac{35}{63}-\dfrac{63}{63}=\dfrac{8}{63}$
4. 豆豆和哥哥从超市买完水果出来，哥哥手里拿着$\dfrac{8}{5}$千克荔枝，豆豆手里拿着一些蓝莓。如果哥哥给豆豆$\dfrac{3}{8}$千克荔枝，那么两人手里的水果一样重。两人一共买了多少千克水果？
答案：4. $\dfrac{8}{5}-\dfrac{3}{8}=\dfrac{49}{40}$（千克）
$\dfrac{49}{40}+\dfrac{49}{40}=\dfrac{49}{20}$（千克）
$\dfrac{49}{40}+\dfrac{49}{40}=\dfrac{49}{20}$（千克）
5. 把15千克钉子分装在三个相同的空木箱里，第一箱连箱重$\dfrac{14}{3}$千克，第二箱连箱重$\dfrac{35}{6}$千克，第三箱连箱重$\dfrac{15}{2}$千克。一个空木箱有多重？
答案：5. $\dfrac{14}{3}+\dfrac{35}{6}+\dfrac{15}{2}=18$（千克）
$18-15=3$（千克） $3÷3=1$（千克）
$18-15=3$（千克） $3÷3=1$（千克）
计算：
$\dfrac{1}{1× 2}+\dfrac{2}{1× 2× 3}+\dfrac{3}{1× 2× 3× 4}+\dfrac{4}{1× 2× 3× 4× 5}+\dfrac{5}{1× 2× 3× 4× 5× 6}+\dfrac{6}{1× 2× 3× 4× 5× 6× 7}=$(
$\dfrac{1}{1× 2}+\dfrac{2}{1× 2× 3}+\dfrac{3}{1× 2× 3× 4}+\dfrac{4}{1× 2× 3× 4× 5}+\dfrac{5}{1× 2× 3× 4× 5× 6}+\dfrac{6}{1× 2× 3× 4× 5× 6× 7}=$(
$\dfrac{5039}{5040}$
)答案：$\dfrac{5039}{5040}$ 提示：因为$\dfrac{a-b}{a× b}=\dfrac{a}{a× b}-\dfrac{b}{a× b}$，通过观察可知，$\dfrac{2}{1×2×3}=\dfrac{3-1}{1×2×3}=\dfrac{3}{1×2×3}-\dfrac{1}{1×2×3}=\dfrac{1}{1×2}-\dfrac{1}{1×2×3}$，同理，$\dfrac{3}{1×2×3×4}=\dfrac{4-1}{1×2×3×4}=\dfrac{1}{1×2×3}-\dfrac{1}{1×2×3×4}$……所以$\dfrac{1}{1×2}+\dfrac{2}{1×2×3}+\dfrac{3}{1×2×3×4}+\dfrac{4}{1×2×3×4×5}+\dfrac{5}{1×2×3×4×5×6}+\dfrac{6}{1×2×3×4×5×6×7}=\dfrac{1}{1×2}+\dfrac{3-1}{1×2×3}+\dfrac{4-1}{1×2×3×4}+\dfrac{5-1}{1×2×3×4×5}+\dfrac{6-1}{1×2×3×4×5×6}+\dfrac{7-1}{1×2×3×4×5×6×7}=\dfrac{1}{1×2}+\dfrac{1}{1×2}-\dfrac{1}{1×2×3}+\dfrac{1}{1×2×3}-\dfrac{1}{1×2×3×4}+···+\dfrac{1}{1×2×3×4×5×6}-\dfrac{1}{1×2×3×4×5×6×7}=1-\dfrac{1}{5040}=\dfrac{5039}{5040}$。
解析：
$\dfrac{1}{1×2}+\dfrac{2}{1×2×3}+\dfrac{3}{1×2×3×4}+\dfrac{4}{1×2×3×4×5}+\dfrac{5}{1×2×3×4×5×6}+\dfrac{6}{1×2×3×4×5×6×7}$
$=\dfrac{1}{1×2}+\dfrac{3-1}{1×2×3}+\dfrac{4-1}{1×2×3×4}+\dfrac{5-1}{1×2×3×4×5}+\dfrac{6-1}{1×2×3×4×5×6}+\dfrac{7-1}{1×2×3×4×5×6×7}$
$=\dfrac{1}{1×2}+(\dfrac{3}{1×2×3}-\dfrac{1}{1×2×3})+(\dfrac{4}{1×2×3×4}-\dfrac{1}{1×2×3×4})+(\dfrac{5}{1×2×3×4×5}-\dfrac{1}{1×2×3×4×5})+(\dfrac{6}{1×2×3×4×5×6}-\dfrac{1}{1×2×3×4×5×6})+(\dfrac{7}{1×2×3×4×5×6×7}-\dfrac{1}{1×2×3×4×5×6×7})$
$=\dfrac{1}{1×2}+(\dfrac{1}{1×2}-\dfrac{1}{1×2×3})+(\dfrac{1}{1×2×3}-\dfrac{1}{1×2×3×4})+(\dfrac{1}{1×2×3×4}-\dfrac{1}{1×2×3×4×5})+(\dfrac{1}{1×2×3×4×5}-\dfrac{1}{1×2×3×4×5×6})+(\dfrac{1}{1×2×3×4×5×6}-\dfrac{1}{1×2×3×4×5×6×7})$
$=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{5040}$
$=1-\dfrac{1}{5040}$
$=\dfrac{5039}{5040}$
$\dfrac{5039}{5040}$
$=\dfrac{1}{1×2}+\dfrac{3-1}{1×2×3}+\dfrac{4-1}{1×2×3×4}+\dfrac{5-1}{1×2×3×4×5}+\dfrac{6-1}{1×2×3×4×5×6}+\dfrac{7-1}{1×2×3×4×5×6×7}$
$=\dfrac{1}{1×2}+(\dfrac{3}{1×2×3}-\dfrac{1}{1×2×3})+(\dfrac{4}{1×2×3×4}-\dfrac{1}{1×2×3×4})+(\dfrac{5}{1×2×3×4×5}-\dfrac{1}{1×2×3×4×5})+(\dfrac{6}{1×2×3×4×5×6}-\dfrac{1}{1×2×3×4×5×6})+(\dfrac{7}{1×2×3×4×5×6×7}-\dfrac{1}{1×2×3×4×5×6×7})$
$=\dfrac{1}{1×2}+(\dfrac{1}{1×2}-\dfrac{1}{1×2×3})+(\dfrac{1}{1×2×3}-\dfrac{1}{1×2×3×4})+(\dfrac{1}{1×2×3×4}-\dfrac{1}{1×2×3×4×5})+(\dfrac{1}{1×2×3×4×5}-\dfrac{1}{1×2×3×4×5×6})+(\dfrac{1}{1×2×3×4×5×6}-\dfrac{1}{1×2×3×4×5×6×7})$
$=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{5040}$
$=1-\dfrac{1}{5040}$
$=\dfrac{5039}{5040}$
$\dfrac{5039}{5040}$