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29 tháng 3 2018

Đặt :

\(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-........-\dfrac{1}{3.1}\)

\(=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+........+\dfrac{1}{3.1}\right)\)

\(=\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.......+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)

\(=-\dfrac{4751}{9603}\)

30 tháng 9 2023

`#3107.101107`

\(B=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{97\cdot99}\right)-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{9603}-\dfrac{1}{2}\cdot\dfrac{96}{97}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{9603}-\dfrac{96}{97}\right)\\ =\dfrac{1}{2}\cdot\left(-\dfrac{9502}{9603}\right)\\ =-\dfrac{4751}{9603}\)

Vậy, `B = -4751/9603.`

30 tháng 9 2023

\(B=\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(B=\dfrac{1}{97.99}-\left(\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\right)\)

Đặt \(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)

\(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)

\(C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{95.97}\)

\(C=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\right):2\)

\(2C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\)

\(2C=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5} +...+\dfrac{1}{95}-\dfrac{1}{97}\)

\(2C=\dfrac{1}{1}-\dfrac{1}{97}\)

\(2C=\dfrac{96}{97}\)

\(C=\dfrac{96}{97}:2=\dfrac{48}{97}\)

Thay C vào ta được:

\(B=\dfrac{1}{97.99}-\dfrac{48}{97}\)

\(99B=\dfrac{99}{97.99}-\dfrac{48.99}{97}\)

\(99B=\dfrac{1}{97}-\dfrac{4752}{97}\)

\(99B=-\dfrac{4751}{97}\)

\(B=-\dfrac{4751}{97}:99=-\dfrac{4751}{9603}\)

20 tháng 9 2023

\(B=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-\dfrac{1}{95\cdot93}-...-\dfrac{1}{3\cdot1}\)

\(B=-\left(\dfrac{1}{3\cdot1}+\dfrac{1}{5\cdot3}+...+\dfrac{1}{97\cdot99}\right)\)

\(2B=-\left(\dfrac{2}{3\cdot1}+\dfrac{2}{5\cdot3}+...+\dfrac{2}{99\cdot97}\right)\)

\(2B=-\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(2B=-\left(1-\dfrac{1}{99}\right)\)

\(2B=-\dfrac{98}{99}\)

\(B=-\dfrac{98}{198}\)

Cậu ơi, \(\dfrac{1}{99\cdot97}\) là dương mà sao lại đưa vào ngoặc âm tất cả vậy nhỉ?

3 tháng 4 2017

Ta có \(A=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{3\cdot1}\)

\(\Leftrightarrow2A=\dfrac{2}{99\cdot97}-\dfrac{2}{97\cdot95}-...-\dfrac{2}{3\cdot1}\)

\(=-\dfrac{1}{99}+\dfrac{1}{97}-\dfrac{1}{97}+\dfrac{1}{95}-...-\dfrac{1}{3}+1\)

\(=-\dfrac{1}{99}+1=\dfrac{98}{99}\)

\(\Rightarrow A=\dfrac{49}{99}\)

3 tháng 4 2017

\(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)

Đặt \(B=\dfrac{1}{97.95}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\)

\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{95.97}\)

\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\)

\(2B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{95}-\dfrac{1}{97}\)

\(2B=1-\dfrac{1}{97}\)

\(2B=\dfrac{96}{97}\)

\(B=\dfrac{96}{97}:2=\dfrac{48}{97}\)

\(\Rightarrow A=\dfrac{1}{99.97}-B=\dfrac{1}{9603}-\dfrac{48}{97}=\dfrac{-4751}{9603}\)

13 tháng 3 2017

Mình sửa lại chút.

\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(=\dfrac{1}{99.97}-\left\{\dfrac{1}{97.95}+\dfrac{1}{95.93}\right\}-\left\{\dfrac{1}{5.3}+\dfrac{1}{3.1}\right\}\)

\(=\dfrac{1}{99.97}-\dfrac{1}{95}.\left\{\dfrac{1}{97}+\dfrac{1}{93}\right\}-\dfrac{1}{3}.\left\{\dfrac{1}{5}+\dfrac{1}{1}\right\}\)

\(=\dfrac{1}{99.97}-\dfrac{1}{95}.\dfrac{190}{97.93}-\dfrac{1}{3}.\dfrac{6}{5}\)

\(=\dfrac{1}{99.97}-\dfrac{2}{97.93}-\dfrac{6}{15}\)

\(=\dfrac{1}{97}.\left\{\dfrac{1}{99}-\dfrac{2}{93}\right\}-\dfrac{2}{5}\)

\(=\dfrac{-35}{297693}-\dfrac{2}{5}\)

\(=\dfrac{-175-595386}{1488465}\)

\(=\dfrac{-595561}{1488465}\)

12 tháng 3 2017

Tách ra và rút gọn là xong bạn nhé !!

a: =11/7(-3/7+4/11-4/7+7/11)=0

b: \(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=-\dfrac{4751}{9603}\)

20 tháng 3 2017

Ta có:

\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)

\(=\dfrac{1}{99.97}=\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)

\(=\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)

\(=\dfrac{1}{9603}-\dfrac{48}{97}\)

\(=\dfrac{-4751}{9603}\)

Vậy \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}=\dfrac{-4751}{9603}\)

11 tháng 3 2017

\(T=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\)

\(T=\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)

Đặt \(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\)

\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)

\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(A=\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)=\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{48}{97}\)

Thay \(A\) vào \(T\) ta có:\(T=\dfrac{1}{99\cdot97}-\dfrac{48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)

11 tháng 3 2017

Đặt \(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(A=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)

\(A=\dfrac{1}{99.97}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\right)\)

Đặt \(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\)

\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}\)

\(2B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\)

\(2B=1-\dfrac{1}{97}\)

\(2B=\dfrac{96}{97}\)

\(B=\dfrac{96}{97}:2\)

\(B=\dfrac{48}{97}\)

\(\Rightarrow A=\dfrac{1}{99.97}-\dfrac{48}{97}\)

\(A=\dfrac{1}{99.97}-\dfrac{48.99}{97.99}\)

\(A=\dfrac{1-48.99}{99.97}\)

\(A=-\dfrac{4751}{9603}\)

Vậy \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}=-\dfrac{4751}{9603}\)

17 tháng 3 2017

K bit có đúng k nhưng cứ nói thử k/q =-6148/15345

17 tháng 3 2017

\(\dfrac{-4751}{9603}\)