\(\frac{1}{12}\)+\(\frac{1}{20}\)+1/30+1/42+1/56+1/72+1/90+1/110+1/132

=\(\frac{1}{3\cdot4}\)+\(\frac{1}{4.5}\)+1/5x6+1/6x7+1/7x8+1/8x9+...1/11x12

=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+1/11-1/12

=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+1/11--1/12

=1/3-1/12

=1/4

Xin lỗi bạn nhé!vì trưa rồi nên mình làm vậy cho nhanh thôi!hjhj!

Nếu thấy mình làm đúng thì k mình nha!Thanks các bạn nhìu!

=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+1/11+1/12

=1/3-1/12

=4/12-1/12

=3/12

=1/4

Kết quả = 

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

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

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

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

= 9/52

dễ ợt như thế mà ko biết làm à 

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

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

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

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

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

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

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

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

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

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

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 = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

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

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

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

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

Hình như đề thiếu

Toi dau co lam gi sai dau, khung ha?

A = 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 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

Vậy A = 7/60.

Xét A , ta thấy:

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

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

Ta lại thấy: \(\frac{1}{5.6}=\frac{1}{5}-\frac{1}{6}\)

\(\frac{1}{6.7}=\frac{1}{6}-\frac{1}{7}\)

....................

\(\frac{1}{11.12}=\frac{1}{11}-\frac{1}{12}\)

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

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

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

\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

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

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

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

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

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