\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+..........+\frac{1}{132}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{11.12}\)

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

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

\(=\frac{11}{12}\)

bài này hóc búa thật

hơi khó 5 chút

=1/1*2+1/2*3+1/3*4+...+1*10*11+1/11*12=1-1/2+1/2-1/3+1/3-1/4+...+1/10-1/11+1/11-1/12

=1-1/12=11/12.

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}+\frac{1}{132}\)

\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{10\times11}+\frac{1}{11\times12}\)

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

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

\(=\frac{11}{12}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}+\frac{1}{132}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}+\frac{1}{11.12}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

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

\(=\frac{11}{12}\)

k mình nha ! Chúc bạn học giỏi ! ^_^

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

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

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

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

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

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{11\cdot12}\)

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

\(=\frac{1}{2}-\frac{1}{12}\)\(=\frac{6}{12}-\frac{1}{12}=\frac{5}{12}\)

1/2 + 1/6 + 1/12 + ... + 1/132

= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/11.12

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/11 - 1/12

= 1 - 1/12

= 11/12

1/90 + 1/110 + 1/132 + ... + 1/10100

= 1/9.10 + 1/10.11 + 1/11.12 + ... + 1/100.101

= ... [như trên]

= 1/9 - 1/100

= 49/450

balabal;a

A = 1 - \(\dfrac{5}{6}\)+\(\dfrac{7}{12}\)-\(\dfrac{9}{20}\)+\(\dfrac{11}{30}\)-\(\dfrac{13}{42}\)+\(\dfrac{15}{56}\) - \(\dfrac{17}{72}\)+\(\dfrac{19}{90}\)+\(\dfrac{23}{132}\)-\(\dfrac{25}{156}\)

A = 1 - \(\dfrac{5}{2.3}\)+\(\dfrac{7}{3.4}\)-\(\dfrac{9}{4.5}\)+\(\dfrac{11}{5.6}\)-\(\dfrac{13}{6.7}\)+\(\dfrac{15}{7.8}\)-\(\dfrac{17}{8.9}\)+\(\dfrac{19}{9.10}\)+\(\dfrac{23}{11.12}\)-\(\dfrac{25}{12.13}\)

A = 1 - \(\dfrac{1}{2}-\dfrac{1}{3}\)+\(\dfrac{1}{3}+\dfrac{1}{4}\)-\(\dfrac{1}{4}-\dfrac{1}{5}\)+...+\(\dfrac{1}{11}+\dfrac{1}{12}\)- \(\dfrac{1}{12}-\dfrac{1}{13}\)

A = 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{13}\) 

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{13}\)

A = \(\dfrac{11}{26}\) 

B = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + ... + 1/110

B = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + .... + 1/10.110

B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/10 - 1/11

B = 1 - 1/11

B = 10/11

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)

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

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

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)

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

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

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

\(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}\)

\(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{4}-\frac{1}{12}=\frac{3-1}{12}=\frac{2}{12}=\frac{1}{6}\)

\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{110}+\frac{1}{132}\)
\(\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+\frac{1}{8\times9}+...+\frac{1}{11\times12}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{4}-\frac{1}{12}=\frac{3-1}{12}\)
\(=\frac{2}{12}=\frac{1}{6}\)