\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)

\(A=\frac{1}{6}-\frac{1}{15}\)

\(A=\frac{1}{10}\)

A=\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

=\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

=\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{14}-\frac{1}{15}\)

=\(\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+...+\frac{1}{210}=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+...+\frac{1}{14.15}\)

\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{14}-\frac{1}{15}\)

\(=\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)

\(A=\frac{1}{6}-\frac{1}{15}\)

\(A=\frac{1}{10}\)

A = 1/5 + [1/5.6 + 1/6.7 + ... + 1/12.13]

A = 1/5 + [1/5-1/6+1/6-1/7+...+1/12-1/13]

A = 1/5 + [1/5-1/13]

A = 1/5 + 8/65

A = 21/65

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{12.13}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+..+\frac{1}{12}-\frac{1}{13}\)

\(=\frac{1}{4}-\frac{1}{13}=\frac{9}{52}\)

= 9/52

dễ ợt như thế mà ko biết làm à 

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)

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

\(A=\frac{1}{5}-\frac{1}{12}\)

\(A=\frac{7}{60}\)

A = \(\frac{1}{5.6}+\frac{1}{6.7}+...+\)\(\frac{1}{10.11}+\frac{1}{11.12}\)

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

A = \(\frac{1}{5}-\frac{1}{12}\)

A = \(\frac{7}{60}\)

dễ đúng không

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+.....+\frac{1}{132}\)

\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.....+\frac{1}{11.12}\)

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

\(A=\frac{1}{5}-\frac{1}{12}\)

\(A=\frac{7}{60}\)

Ta có:

A = \(\frac{1}{5.6}\)+ \(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)+\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)

A bằng 7/60

Bạn xem lời giải của mình nhé:

Giải:

\(A=\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\\ =\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\\ =\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\\ =\frac{1}{5}-\frac{1}{12}=\frac{12-5}{60}=\frac{7}{60}\)

Chúc bạn học tốt!

A=\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
  =1/5 - 1/6+1/6 -1/7+1/7-1/8+....+1/11 - 1/12
  = 1/5 - 1/12
  =12/60 -5/60
  = 7/60

cảm ơn bn nha

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

\(A=\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{11\times12}\)

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

\(A=\frac{1}{4}-\frac{1}{12}\)

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