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

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

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

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

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​

A bn lướt xuống dưới mà xem cách làm 

nhưng của bn là cho 3 ra ngoài nha

Giải:

A=3/1.3+3/3.5+3/5.7+...+3/49.51

A=3/2.(2/1.3+2/3.5+2/5.7+...+2/49.51)

A=3/2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/49-1/51)

A=3/2.(1/1-1/51)

A=3/2.50/51

A=25/17

B=1/3+1/3²+1/3³+...+1/3⁸

3B=1+1/3+1/3²+...+1/3⁷

3B-B=(1+1/3+1/3²+...+1/3⁷)-(1/3+1/3²+1/3³+...+1/3⁸)

2B=1-1/3⁸

   B=1-1/3⁸ /2

Chúc bạn học tốt!

\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)

\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)

\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)

=168/101

\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\)                                                                                                          \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(B=2.\left(\frac{1}{1}-\frac{1}{101}\right)\)

\(B=2.\frac{100}{101}=\frac{200}{101}\)

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

Ta có :

M= \(\dfrac{3+3-3+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4+4-4+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)= \(\dfrac{3+3-3}{4+4-4}=\dfrac{3}{4}\)

b) Nhận xét thấy: \(\dfrac{2}{1.3}=1-\dfrac{1}{3};\dfrac{1}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};...\)

Ta có:

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

B= 1- \(\dfrac{1}{101}\)= \(\dfrac{100}{101}\)

Vậy B= \(\dfrac{100}{101}\)

cho minh xin yeu cau de bai

trả hiểu yêu cầu đề bài là j cả

\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\\ =\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =2.\left(1-\dfrac{1}{101}\right)\\ =2.\dfrac{100}{101}\\ =\dfrac{200}{101}\)

`4/1.3+4/3.5+4/5.7+...+4/99.101`

`=2(2/1.3+2/3.5+2/5.7+...+2/99.101)`

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

`=2(1-1/101)`

`=2. 100/101`

`=200/101`

a: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{202}{203}\)

b: \(=-4\left(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{2015\cdot2018}\right)\)

\(=-\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{2015\cdot2018}\right)\)

\(=\dfrac{-4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{2015}-\dfrac{1}{2018}\right)\)

\(=\dfrac{-4}{3}\cdot\dfrac{504}{1009}=-\dfrac{672}{1009}\)