1/2=1/1.2

1/6=1/2.3

1/12=1/3.4

1/20=1/4.5

1/30=1/5.6

1/42=1/6.7

ta có 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

= 1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

=1/1-1/7

=6/7

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{7}\)

=\(1-\frac{1}{7}\)=\(\frac{6}{7}\)

ta có:

1/6+1/12+1/20+1/30+.........+1/90+1/110

= 1/2x3+1/3x4+1/4x5+1/5x6+....+1/9x10+1/10x11

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

=1/2-1/11=11/22-2/22=9/22

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{10}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{11}=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)

M  =  \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

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

M     =   1   -\(\frac{1}{9}\)=\(\frac{8}{9}\)

trả lời: 8/9

M=1/2+1/6+1/12+1/20+.....+1/72

    =1/(1.2)+1/(2.3)+1/(3.4)+....+1/(8.9)

    =1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9

    =1-1/9=8-9

1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2

= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)

= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)

= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)

= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)

= 1/90 - ( 1 - 1/9)

= 1/90 - 8/9

= 1/90 - 80/90

= -79/90

1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2

= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)

= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)

= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)

= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)

= 1/90 - ( 1 - 1/9)

= 1/90 - 8/9

= 1/90 - 80/90

= -79/90

 mk nha cac ban 

ta nhận thấy

1/2=1-1/2

1/6=1/2-1/3

1/12=1/3-1/4

1/20=1/4-1/5

1/30=1/5-1/6

1/42=1/6-1/7

ta có:

1/2+1/6+1/12+1/20+1/30+1/42

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

=1-1/7

=6/7

bn tự hiểu nha

1/2+1/6+1/12+1/20+1/30+1/42 = \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\) = 1-1/7=6/7

\(\frac{1}{2}\)+ \(\frac{1}{6}+\frac{1}{12}+\)\(\frac{1}{20}\)+ \(\frac{1}{30}\)+ \(\frac{1}{42}\)

= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)

= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

= \(\frac{1}{1}-\frac{1}{7}\)

= \(\frac{7}{7}-\frac{1}{7}\)

= \(\frac{6}{7}\)

Bài làm 

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

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

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

\(A=\frac{10}{11}\)

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

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

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