\(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+...+\frac{1}{440}\)

\(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}+....+\frac{1}{20\cdot22}\)

\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+.....+\frac{2}{20\cdot22}\)

\(2A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{20}-\frac{1}{22}\)

\(2A=1-\frac{1}{22}\)

\(A=\frac{21}{22}:2\)

\(A=\frac{21}{44}\)

\(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+...+\frac{1}{440}\)

= \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{20.22}\)

= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{20}-\frac{1}{22}\right)\)

= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{22}\right)=\frac{1}{2}.\frac{5}{11}=\frac{5}{22}\)

\(S=\frac{1}{8}+\frac{1}{24}+\frac{1}{80}+...+\frac{1}{440}\)

=> \(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+...+\frac{1}{20.22}\)

=> \(S=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{20.22}\right)\)

=> \(S=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)

=>\(S=\frac{1}{2}.\left(1-\frac{1}{11}\right)=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

Mk làm lại nhé sorry bn lúc nãy mk nhầm

\(D=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+........+\frac{1}{1520}\)

\(D=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+........+\frac{1}{38.40}\)

\(2D=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+......+\frac{2}{38.40}\)

\(2D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-.......-\frac{1}{40}\)

\(2D=1-\frac{1}{40}\)

\(2D=\frac{40}{40}-\frac{1}{40}\)

\(2D=\frac{39}{40}\)

\(D=\frac{39}{40}:2=\frac{39}{40}.\frac{1}{2}=\frac{39}{80}\)

Vậy ....

\(D=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{1520}\)

\(D=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{38.40}\)

\(D=\frac{1}{2}\times\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{38.40}\right)\)

\(D=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{38}-\frac{1}{40}\right)\)

\(D=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{40}\right)\)

\(D=\frac{1}{2}\times\frac{19}{40}\)

\(D=\frac{19}{80}\)

_Chúc bạn học tốt_

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

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

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

\(=1+\dfrac{1}{2}-\dfrac{1}{12}=\dfrac{17}{12}\)

\(1\frac{5}{24}=\frac{29}{24}\)

A = 1 + 1/2.4 + 1/4.6 + 1/6.8 + 1/8.10 + 1/10.12

2A = 2 + 2/2.4 + 2/4.6 + 2/6.8 + 2/8.10 + 2/10.12

     = 2 + 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + 1/8 - 1/10 + 1/10 - 1/12

     = 2 + 1/2 - 1/12 = 29/12

=> A = 29/12 : 2 = 29/24

Tk mk nha

A =1+ 1/2.4 + 1/4.6 +1/6.8+1/8.10+1/10.12

=1+1/2 -1/4 +1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12

=1+1/2-1/12

=3/2-1/12

=17/12

A=1+18+124+148+180+1120A=1+18+124+148+180+1120

=1+12.4+14.6+16.8+18.10+110.12=1+12.4+14.6+16.8+18.10+110.12

=1+12(12−14+14−16+16−18+18−110+110−112)=1+12(12−14+14−16+16−18+18−110+110−112)

=1+12(12−112)=1+12(12−112)

=1+524=1+524

=2924

Tham khảo thôi nka

2A= 2/8+2/24+2/48+2/80= 2/(2*4)+2/(4*6)+2/(6*8)+2/(8*10)= 1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10= 1/2-1/10= 2/5 =>A= 1/5

\(A=\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}=\dfrac{1}{2\times4}+\dfrac{1}{4\times6}+\dfrac{1}{6\times8}+\dfrac{1}{8\times10}\)

\(2A=\dfrac{2}{2\times4}+\dfrac{2}{4\times6}+\dfrac{2}{6\times8}+\dfrac{2}{8\times10}\\ 2A=\dfrac{2}{2}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{6}+\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{8}-\dfrac{2}{10}\\ 2A=\dfrac{2}{2}-\dfrac{2}{10}=\dfrac{4}{5}\\ A=\dfrac{2}{5}\)

`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 \(\dfrac{5}{12}\)

= \(\dfrac{5}{24}\)

49/48

\(A=\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{35}+\frac{1}{48}+\frac{1}{63}+\frac{1}{80}\)

\(=\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\frac{1}{4\times6}+\frac{1}{5\times7}+\frac{1}{6\times8}+\frac{1}{7\times9}+\frac{1}{8\times10}\)

\(=\frac{1}{2}\times\left[\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\right)+\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\frac{2}{8\times10}\right)\right]\)

\(=\frac{1}{2}\times\left[\left(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}\right)+\left(\frac{4-2}{2\times4}+\frac{6-4}{4\times6}+\frac{8-6}{6\times8}+\frac{10-8}{8\times10}\right)\right]\)

\(=\frac{1}{2}\times\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\right]\)

\(=\frac{1}{2}\times\left[\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)

\(=\frac{29}{45}\)

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 \(A=\dfrac{29}{24}\)