10. 计算：
(1) $(+6)-(-3)+(-2)-(+6)$.
(2) $9-(-0.3)+(-6)-(-4.7)$.
(3) $\frac{2}{9}+1\frac{5}{6}-(-\frac{2}{9})-\frac{1}{2}$.
(4) $(+\frac{1}{3})-(+\frac{5}{6})+(-\frac{1}{6})-(-\frac{2}{3})$.

10. (1) 原式$=6+3-2-6=1$. (2) 原式$=9+0.3-6+4.7=(9-6)+(0.3+4.7)=3+5=8$. (3) 原式$=\dfrac{2}{9}+1\dfrac{5}{6}+\dfrac{2}{9}-\dfrac{1}{2}=\dfrac{16}{9}$. (4) 原式$=\dfrac{1}{3}-\dfrac{5}{6}-\dfrac{1}{6}+\dfrac{2}{3}=(\dfrac{1}{3}+\dfrac{2}{3})-(\dfrac{5}{6}+\dfrac{1}{6})=1-1=0$.

11. 教材P42习题T10变式 某中学开展一分钟跳绳比赛，成绩以200下为标准数量，超过的数量记为正数，不足的数量记为负数. 七年级某班的8名同学组成代表队参赛，成绩情况记录如下(单位：下)：+8,0,−5,+12,−9,+1,+8,+15.
（1）该班参赛代表队中的最好成绩与最差成绩相差多少下？
（2）该班参赛代表队一共跳了多少下？
（3）规定：每分钟跳绳数量为标准数量，不得分；超过标准数量，每多跳1下得2分；未达到标准数量，每少跳1下扣1分. 若参赛代表队跳绳总得分超过70分，便可得到学校的奖励，请通过计算说明该班参赛代表队能否得到学校的奖励.

11. (1) $+15-(-9)=15+9=24$(下).所以该班参赛代表队中的最好成绩与最差成绩相差24下. (2) $200×8+(+8)+0+(-5)+(+12)+(-9)+(+1)+(+8)+(+15)=1\ 630$(下).所以该班参赛代表队一共跳了1 630下. (3) $(8+12+1+8+15)×2-(5+9)×1=74$(分).因为$74＞70$,所以该班参赛代表队能得到学校的奖励.

1. [简便运算]计算：$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}=$.

1. $\dfrac{2}{5}$ 原式$=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}$.

2. [简便运算]计算：$\frac{5}{6}+\frac{23}{24}+\frac{59}{60}+\frac{83}{84}+\frac{119}{120}+\frac{209}{210}-6=$。

2. $-\dfrac{1}{4}$ 因为$\dfrac{5}{6}=1-\dfrac{1}{6}$,$\dfrac{23}{24}=1-\dfrac{1}{24}$,$\dfrac{59}{60}=1-\dfrac{1}{60}$,$\dfrac{83}{84}=1-\dfrac{1}{84}$,$\dfrac{119}{120}=1-\dfrac{1}{120}$,$\dfrac{209}{210}=1-\dfrac{1}{210}$,又因为$\dfrac{1}{6}=\dfrac{1}{1×2×3}$,$\dfrac{1}{24}=\dfrac{1}{2×3×4}$,$\dfrac{1}{60}=\dfrac{1}{3×4×5}$,$\dfrac{1}{84}=\dfrac{1}{3×4×7}$,$\dfrac{1}{120}=\dfrac{1}{4×5×6}$,$\dfrac{1}{210}=\dfrac{1}{5×6×7}$,所以原式$=6-(\dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+\dfrac{1}{84}+\dfrac{1}{120}+\dfrac{1}{210})-6=-(\dfrac{1}{1×2×3}+\dfrac{1}{2×3×4}+\dfrac{1}{3×4×5}+\dfrac{1}{3×4×7}+\dfrac{1}{4×5×6}+\dfrac{1}{5×6×7})=-[\dfrac{1}{2}×(\dfrac{1}{1×2}-\dfrac{1}{2×3})+\dfrac{1}{2}×(\dfrac{1}{2×3}-\dfrac{1}{3×4})+\dfrac{1}{2}×(\dfrac{1}{3×4}-\dfrac{1}{4×5})+\dfrac{1}{3×4×7}+\dfrac{1}{2}×(\dfrac{1}{4×5}-\dfrac{1}{5×6})+\dfrac{1}{2}×(\dfrac{1}{5×6}-\dfrac{1}{6×7})]=-\dfrac{1}{2}×(\dfrac{1}{1×2}-\dfrac{1}{2×3}+\dfrac{1}{2×3}-\dfrac{1}{3×4}+\dfrac{1}{3×4}-\dfrac{1}{4×5}+\dfrac{1}{4×5}-\dfrac{1}{5×6}+\dfrac{1}{5×6}-\dfrac{1}{6×7})-\dfrac{1}{3×4×7}=-\dfrac{1}{2}×(\dfrac{1}{1×2}-\dfrac{1}{6×7})-\dfrac{1}{3×4×7}=-\dfrac{5}{21}-\dfrac{1}{84}=-\dfrac{1}{4}$.