= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)

 = \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)

 = \(\dfrac{5}{2}.\dfrac{100}{101}\)

 = \(\dfrac{250}{101}\)

\(=5\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\right)\)

\(=5\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)\)

\(=5\cdot\dfrac{5}{24}=\dfrac{25}{24}\)

`5/12 + 5/20 + 5/30 + 5/42 + 5/56`

`= 5/(3.4) + 5/(4.5) + 5/(5.6) + 5/(6.7) + 5/(7.8)`

`= 5 . (1/(3.4) + 1/(4.5) + ... + 5/(7.8))`

`= 5 . (1/3 - 1/4 + 1/4 - 1/5 + ...+ 1/7 - 1/8)`

`= 5 . (1/3 - 1/8) `

`= 5 . 5/24 = 25/24`

\(=1-\dfrac{1}{2}+1-\dfrac{1}{6}+...+1-\dfrac{1}{90}\)

\(=10-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)

\(=10-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

=9+1/10

=9,1

\(A=\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}\\ =\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+\left(1-\dfrac{1}{30}\right)+\left(1-\dfrac{1}{42}\right)+\left(1-\dfrac{1}{56}\right)\\ =6-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\right)\\ =6-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}\right)\\ 6-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\right)\\ =6-\left(1-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{8}\right)=\dfrac{207}{40}\)

Chắc đề thiếu, nhưng mình cứ làm hướng kiểu kia trước nhaa

Giải:

1/2+ 5/6+ 11/12+ 29/30+ 41/42+ 55/56

=(1- 1/2)+ (1-1/6)+ (1- 1/12)+ (1-1/30)+ (1- 1/42)+ (1- 1/56)

=1- 1/2+ 1- 1/6+1-1/12 1-1/30+1-1/42+1-1/56

=6-(1/2+1/6+1-1/12+1/30+1/42+1/56)

=6-(1/1.2+1/2.3+1/3.4+1/5.6+1/6.7+1/7.8)

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

=6-(1/1-1/4+1/5-1/8)

=6-33/40

=207/40

Chúc bạn học tốt!

\(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}\)

\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{5}{6}\right)+\left(1-\dfrac{11}{12}\right)+\left(1-\dfrac{29}{30}\right)+\left(1-\dfrac{41}{42}\right)\left(1-\dfrac{55}{56}\right)\)

\(=6-\left(\dfrac{1}{1.2}+\dfrac{5}{2.3}+\dfrac{11}{3.4}+\dfrac{29}{5.6}+\dfrac{41}{6.7}+\dfrac{55}{7.8}\right)\)

\(=6-\left(1-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{8}\right)=\dfrac{207}{40}\)

8/9

\(=1-\dfrac{1}{2}+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+1-\dfrac{1}{30}+1-\dfrac{1}{42}+1-\dfrac{1}{56}+1-\dfrac{1}{72}+1-\dfrac{1}{90}\)

\(=9-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=9-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(=9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=9-\left(1-\dfrac{1}{10}\right)\)

\(=9-\dfrac{9}{10}\)

\(=\dfrac{81}{10}\)

a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}+\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)

\(=\dfrac{3}{5}+\dfrac{41}{6}\left(11\dfrac{1}{4}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)

\(=\dfrac{3}{5}+\dfrac{41}{6}.2.\dfrac{3}{25}\)

\(=\dfrac{3}{5}+\dfrac{41}{25}\)

\(=\dfrac{15}{25}+\dfrac{41}{25}\)

\(=\dfrac{56}{25}\)

a) 

b) \(B=\dfrac{-1}{2}+\dfrac{-1}{6}+\dfrac{-1}{12}+\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\) \(1-\dfrac{1}{10}\)

\(=\dfrac{-9}{10}\)

B= \(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{9}{20}+\dfrac{11}{30}-\dfrac{13}{42}+\dfrac{15}{56}\)

= \(\left(1+\dfrac{1}{2}\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)-\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)-\left(\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{7}+\dfrac{1}{8}\right)\)

= 1+\(\dfrac{1}{8}\)=\(\dfrac{9}{8}\)