\(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

\(=\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+\dfrac{-1}{7.8}+\dfrac{-1}{8.9}+\dfrac{-1}{9.10}\)

\(=\dfrac{-1}{4}-\dfrac{-1}{5}+\dfrac{-1}{5}-\dfrac{-1}{6}+\dfrac{-1}{6}-\dfrac{-1}{7}+\dfrac{-1}{7}-\dfrac{-1}{8}+\dfrac{-1}{8}-\dfrac{-1}{9}+\dfrac{-1}{9}-\dfrac{-1}{10}\)

\(=\dfrac{-1}{4}-\dfrac{-1}{10}\)

\(=\dfrac{-3}{20}\)

\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)

\(=\frac{1+2}{1.2}-\frac{2+3}{2.3}+\frac{3+4}{3.4}-\frac{4+5}{4.5}+\frac{5+6}{5.6}-\frac{6+7}{6.7}+\frac{7+8}{7.8}-\frac{8+9}{8.9}+\frac{9+10}{9.10}\)

\(=1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-...-\frac{1}{8}-\frac{1}{9}+\frac{1}{9}+\frac{1}{10}\)

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

các bạn làm hộ tôi nhé

bai nay giai the nao

A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56

A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8

A = 1 + ( -1/2 + 1/2 ) + ( -1/3 + 1/3 ) + ( -1/4 + 1/4 ) + ( -1/5 + 1/5 ) + ( -1/6 + 1/6 ) + ( -1/7 + 1/7 ) - 1/8

A = 1 + 0 + 0 + 0 + 0 + 0 + 0 - 1/8

A = 1 - 1/8

A = 7/8

* Sửa đề tí nhé 

B = 3/2 - 5/6 + 7/12 - 9/20 + 11/30 - 13/42 + 15/56

B = 3/1.2 - 5/2.3 + 7/3.4 - 9/4.5 + 11/5.6 - 13/6.7 + 15/7.8

B = 3 - 3/2 - 5/2 - ( -5/3 ) + 7/3 - 7/4 - 9/4 - ( -9/5 ) + 11/5 - 11/6 - 13/6 - ( -13/7 ) + 15/7 - 15/8

B = 3 - 3/2 - 5/2 + 5/3 + 7/3 - 7/4 - 9/4 + 9/5 + 11/5 - 11/6 - 13/6 + 13/7 + 15/7 - 15/8

B = 3 + ( -3/2 - 5/2 ) + ( 5/3 + 7/3 ) + ( -7/4 - 9/4 ) + ( 9/5 + 11/5 ) + ( -11/6 - 13/6 ) + ( 13/7 + 15/7 ) - 15/8

B = 3 + -4 + 4 + -4 + 4 + -4 + 4 - 15/8

B = 3 + 0 + 0 + 0 - 15/8

B = 3 - 15/8

B = 9/8

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

hơi lệch

\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)

\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)

Bài 1: 

Ta có: \(A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\)

\(=\dfrac{1}{2}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{100}{309}=\dfrac{50}{309}\)

Bài 2: 

Ta có: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)

\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)

\(=7\left(1-\dfrac{1}{10}\right)\)

\(=\dfrac{63}{10}\)

:)? Đề bài là ntn bạn nhỉ??

Bạn ra kết quả rồi mà !!

Tính :

A= 3/2 - 5/6 + 7/12 - 9/20 + 11/30 - 13/42 + 15/56 - 17/72 + 19/70