\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\left(\dfrac{100}{101}\right)=\dfrac{50}{101}\)

\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)

= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ...+ 1/99 + 1/101)

= 1/2 . (1/1 - 1/101)

= 1/2 . 100/101

= 50/101

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{100}{101}=\dfrac{50}{101}\)

A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)

=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)

=2.(1-1/101)

=2.(101/101-1/101)

=2.100/101

200/101

B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)

=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)

=2.(1/1+1/101)

=2.(101/101+1/101)

=2.102/101

=204/101

C=1/2+1/3+1/3+1/4+....+1/99+1/100

=1/2+1/100

=50/100+1/100

=51/100

`A=2/[1.3]+2/[3.5]+2/[5.7]+.....+2/[99.101]`

`A=1-1/3+1/3-1/5+1/5-1/7+......+1/99-1/101`

`A=1-1/101=101-1/101=100/101`

\(\dfrac{100}{101}\)

A=2/1.3+2/3.5+2/5.7+...+2/99.101
   = 1/1 - 1/3 +1/3 - 1/5 +.... +1/99 - 1/101
   = 1-1/101
   =101/101-1/101
   =100/101   

2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)

= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)

=1-\(\dfrac{1}{101}\)=...........

mk làm vậy thôi nha

thông cảm

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)

=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)

tương tự

chữ xấu (thông cảm) ;-;

\(A=1.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{5}+...+\dfrac{1}{2021}.\dfrac{1}{2023}=1-\dfrac{1}{2023}=\dfrac{2022}{2023}\)

\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2021\cdot2023}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)

cho minh xin yeu cau de bai

trả hiểu yêu cầu đề bài là j cả

Đợi mình được không mình làm được câu a

a)\(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+........+\(\dfrac{2}{99.101}\))

=\(\dfrac{1}{2}\).(1-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+........+\(\dfrac{1}{99}\)-\(\dfrac{1}{101}\))

=\(\dfrac{1}{2}\).(1-\(\dfrac{1}{101}\))

=\(\dfrac{1}{2}\).\(\dfrac{100}{101}\)

=\(\dfrac{100}{202}\)=\(\dfrac{50}{101}\)

a) Đặt \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{99.101}\) = A

Ta có:

A = \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{99.101}\)

\(\Rightarrow\) 2A = 2\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{99.101}\right)\)

\(\Rightarrow\) 2A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{99.101}\)

\(\Rightarrow\) 2A = \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{99}-\dfrac{1}{101}\)

\(\Rightarrow\) 2A = 1 - \(\dfrac{1}{101}\) = \(\dfrac{100}{101}\)

\(\Rightarrow\) A = \(\dfrac{100}{101}\) : 2

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2021.2023}\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{2021.2023}\right)\)

\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)

\(=\dfrac{1}{2}.\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}.\dfrac{2022}{2023}=\dfrac{1011}{2023}\)

Ta có A = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2021\cdot2023}\)

            = \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\right)\)

            = \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}+\dfrac{1}{2023}\right)\)

            = \(\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
 

A=1.31​+3.51​+5.71​+.....+2021.20231​

=12.(21.3+23.5+25.7+....+22021.2023)=21​.(1.32​+3.52​+5.72​+....+2021.20232​)

=12.(1−13+13−15+15−17+....+12021−12023)=21​.(1−31​+31​−51​+51​−71​+....+20211​−20231​)

=12.(1−12023)=12.20222023=10112023=21​.(1−20231​)=21​.20232022​=20231011​