\(=\frac{9}{10.11}-\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)

\(=\frac{9}{10.11}-\frac{10-9}{9.10}-\frac{9-8}{8.9}-...-\frac{2-1}{1.2}\)

\(=\frac{9}{10.11}-\frac{10}{9.10}+\frac{9}{9.10}-...-\frac{2}{1.2}+\frac{1}{1.2}\)

\(=\frac{9}{10.11}-\frac{1}{9}+\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-\frac{1}{7}+\frac{1}{8}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)

\(=\frac{9}{10.11}+\frac{1}{10}-1\)

\(=-\frac{9}{11}\)

\(\dfrac{1}{20}=\dfrac{1}{4x5}=\dfrac{1}{4}-\dfrac{1}{5}\)

Tương tự các phân số khác

S= \(\dfrac{1}{4}-\dfrac{1}{12}=\dfrac{1}{6}\)

\(\dfrac{1}{20}+\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)

= \(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)

= \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)

= \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)

= \(\dfrac{3}{12}\) - \(\dfrac{1}{12}\)

= \(\dfrac{2}{12}\)

=\(\dfrac{1}{6}\)

=1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x11+1/11x12

=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12

=1/4-1/12=3/12-1/12=2/12=1/6

`=1/[4xx5]+1/[5xx6]+1/[6xx7]+...+1/[11xx12]`

`=1/4-1/5+1/5-1/6+1/6-1/7+...+1/11-1/12`

`=1/4-1/12=3/12-1/12=2/12=1/6`

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\\ =\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}+\dfrac{1}{11\times12}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{1}{12}\\ =\dfrac{3}{12}-\dfrac{1}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)

ta có: 1/2=1/1x2

          1/6=1/2x3

           1/12=1/3x4

             1/20=1/4x5

             1/30=1/5x6

                1/42=1/6x7

                 1/56=1/7x8

              1/72=1/8x9

               1/90=1/9x10

                1/110=1/10x11

              tiếp theo bn tiếp tục nhé

1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110

=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x1

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11

=1-1/11

=10/11

cảm ơn 2 bạn nhé

a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)

=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)

=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)

=>\(A=\dfrac{127}{128}\)

b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)

1/2+1/6+1/12+...+1/110

=1/1.2+1/2.3+1/3.4+...+1/10.11

=1-1/2+1/2-1/3+1/3-1/4+...+1/10-1/11

=1-1/11=10/11

1/2

\(A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{10}-\frac{1}{10}\right)\)\(=\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)

9/22

**** mọi người thanks nhiều

=1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10+1/10.11                 =1-10/11=9/22    ;minh lanm dung nhe ;nho **** cho minh nhe

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

\(=\frac{1}{2}-\frac{1}{11}\)

\(=\frac{9}{22}\)

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(A=\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0\)

\(A=\frac{11}{22}-\frac{2}{22}\)

\(A=\frac{9}{22}\)

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)

\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)

\(=\dfrac{1}{2}-\dfrac{1}{11}\)

\(=\dfrac{9}{22}\)

\(\Rightarrow\)S\(=\)9/22  nha bạn,mình tính nhần

\(\frac{10}{11}\)nha bạn mình giải rồi

S\(=\)1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11

\(\rightarrow\)S\(=\)1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11

\(\rightarrow\)S\(=\)1/2-1/11

\(\rightarrow\)S\(=\)11/22-2/22

\(\Rightarrow\)S\(=\)7/22