Ta có: \(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)

\(\Leftrightarrow2A=2+\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}\)

\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\)

\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{12}\)

\(\Leftrightarrow2A=\dfrac{24}{12}+\dfrac{6}{12}-\dfrac{1}{12}\)

\(\Leftrightarrow2A=\dfrac{29}{12}\)

hay \(A=\dfrac{29}{24}\)

a: =-21/36-3/36=-24/36=-2/3

b: =43/12*1/2+5/24=43/24+5/24=2

c: =8/9+1/9=1

e: =1-1/4+1/4-1/7+...+1/97-1/100

=1-1/100=99/100

`1/8+1/24+1/48+1/80+1/120`

`=1/[2xx4]+1/[4xx6]+1/[6xx8]+1/[8xx10]+1/[10xx12]`

`=1/2xx(2/[2xx4]+2/[4xx6]+2/[6xx8]+2/[8xx10]+2/[10xx12])`

`=1/2xx(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)`

`=1/2xx(1/2-1/12)`

`=1/2xx(6/12-1/12)`

`=1/2xx5/12=5/24`

\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)

=\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{10.12}\)

=\(\dfrac{1}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{10.12}\right)\)

=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)

=\(\dfrac{1}{2}.\dfrac{5}{12}\)

=\(\dfrac{5}{24}\)

Dấu chấm(.)là nhân.

   \(\dfrac{1}{8}\) + \(\dfrac{1}{24}\)+ \(\dfrac{1}{48}\)+ \(\dfrac{1}{80}\)+ \(\dfrac{1}{120}\)

 = \(\dfrac{1}{2}\) X (\(\dfrac{2}{2\times4}\)+ \(\dfrac{2}{4\times6}\)+\(\dfrac{2}{6\times8}\)+\(\dfrac{2}{8\times10}\)+ \(\dfrac{2}{10\times12}\))

= \(\dfrac{1}{2}\) x ( \(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{12}\))

= \(\dfrac{1}{2}\)x (\(\dfrac{1}{2}\)-\(\dfrac{1}{12}\))

= \(\dfrac{1}{2}\) X \(\dfrac{6-1}{12}\)

= \(\dfrac{1}{2}\)x \(\dfrac{5}{12}\)

= \(\dfrac{5}{24}\)

=\(1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{10\cdot12}\)

\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)

\(=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)

==1+12(12−14+14−16+...+110−112)=1+12(12−14+14−16+...+110−112)

a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)

\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)

b) 

\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)

\(=-14\)

c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)

\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)

\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)

\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)

\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)

a: =-1/3+1/3=0

b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)

c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)

d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)

a: =-1/3+1/3=0

b: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167

d: 

12+3/7-11+3/7 tính bằng cách thuận tiện nhé

\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{30}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{29}{30}\)

\(=\dfrac{1.2.3...29}{2.3.4...30}\)

\(=\dfrac{1}{30}\)

\(B=1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.1\dfrac{1}{24}...1\dfrac{1}{168}\)

\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{169}{168}\)

\(=\dfrac{4.9.16.25...169}{3.8.15.24...168}\)

\(=\dfrac{2.3.4...13}{1.2.3...12}.\dfrac{2.3.4...13}{3.4.5...14}\)

\(=13.\dfrac{1}{7}\)

\(=\dfrac{13}{7}\).

A = \(\dfrac{1}{2}x\dfrac{2}{3}x\dfrac{3}{4}x...x\dfrac{29}{30}=1x1x1x...x\dfrac{1}{30}=\dfrac{1}{30}\)

B = \(\dfrac{4}{3}x\dfrac{9}{8}x\dfrac{16}{15}x\dfrac{25}{24}x...x\dfrac{169}{168}=1x1x1x1x...x\dfrac{13}{7}=\dfrac{13}{7}\)

Câu B em chưa rõ cách làm nhanh cho lắm. Nếu ko cần tính nhanh thì chị có thể giải bình thường ra giấy ha.

\(\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}-\dfrac{47}{24}\right)+\dfrac{2}{3}=\dfrac{37}{37}-\dfrac{13}{8}+\dfrac{2}{3}=\dfrac{1}{24}\)