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

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

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

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

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

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.......+\frac{1}{90}+\frac{1}{110}\)

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

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

\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

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

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+....+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)

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

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

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

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)\(+\frac{1}{110}\) 

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...\) \(+\frac{1}{9\cdot10}\)\(+\frac{1}{10\cdot11}\)

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

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

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

=10/11 đó

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\)

Chỉ cần viết ra là: \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}=1-\frac{1}{11}=\frac{10}{11}\)

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

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

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

1/3-1/12

=1/4

Chúc bạn bạn may mắn!

1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132 = 2/3,

 Đúng 100%!

Ai tk cho mình mình tk lại.

\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{2}-\frac{1}{12}\)

\(=\frac{5}{12}\)

bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại

89/90 Đấy

89/90

Tách mẫu ra như sau:

\(T=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+..\)

Đó cứ vậy nhé