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5 tháng 4 2017

Ta có:

\(A=\dfrac{1}{1.1981}+\dfrac{1}{2.1982}+...+\dfrac{1}{n\left(1980+n\right)}+...+\dfrac{1}{25.2005}\)

\(=\dfrac{1}{1980}\left(\dfrac{1981-1}{1.1981}+\dfrac{1982-2}{2.1982}+...+\dfrac{1980+n-n}{n\left(1980+n\right)}+...+\dfrac{2005-25}{25.2005}\right)\)

\(=\dfrac{1}{1980}\left(1-\dfrac{1}{1981}+\dfrac{1}{2}-\dfrac{1}{1982}+...+\dfrac{1}{n}-\dfrac{1}{1980+n}+...+\dfrac{1}{25}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{1980}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)

Lại có:

\(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{m\left(m+25\right)}+...+\dfrac{1}{1980.2005}\)

\(=\dfrac{1}{25}\left(\dfrac{26-1}{1.26}+\dfrac{27-2}{2.27}+...+\dfrac{25+m-m}{m\left(25+m\right)}+...+\dfrac{2005-1980}{1980.2005}\right)\)

\(=\dfrac{1}{25}\left(\dfrac{1}{1}-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{m}-\dfrac{1}{25+m}+...+\dfrac{1}{1980}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{25}\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{1980}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{2005}\right)\right]\)

\(=\dfrac{1}{25}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}=\dfrac{5}{396}\)

Vậy tỉ số của \(A\)\(B\)\(\dfrac{5}{396}\)

30 tháng 11 2019

này này là bạn j ơi bạn j ơi

này này là cho mk hỏi tẹo dc ko

cái chô bạn rút gọn B ik dòng thứ 4 xuống dòng thứ 5 bạn làm kiểu j vậy giải thích rõ hook mik dc ko thanks nhìu,@Vũ Minh Tuấn@Hoang Hung Quan,@Băng Băng 2k6,

13 tháng 12 2017

giải nè

Image

13 tháng 12 2017

được chưa cho mình k

19 tháng 4 2016

ta có 1980 A=1980

AH
Akai Haruma
Giáo viên
10 tháng 3 2018

Lời giải:

Ta có \(A=\frac{1}{1.1981}+\frac{1}{2.1982}+...+\frac{1}{25.2005}\)

\(\Rightarrow 1980A=\frac{1980}{1.1981}+\frac{1980}{2.1982}+...+\frac{1980}{25.2005}\)

\(\Leftrightarrow 1980A=\frac{1981-1}{1.1981}+\frac{1982-2}{2.1982}+....+\frac{2005-25}{25.2005}\)

\(\Leftrightarrow 1980A=1-\frac{1}{1981}+\frac{1}{2}-\frac{1}{1982}+...+\frac{1}{25}-\frac{1}{2005}\)

\(1980A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)-\left(\frac{1}{1981}+\frac{1}{1982}+..+\frac{1}{2005}\right)\) (1)

Lại có:

\(25B=\frac{25}{1.26}+\frac{25}{2.27}+...+\frac{25}{1980.2005}\)

\(\Leftrightarrow 25B=\frac{26-1}{1.26}+\frac{27-2}{2.27}+...+\frac{2005-1980}{1980.2005}\)

\(\Leftrightarrow 25B=1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+...+\frac{1}{1980}-\frac{1}{2005}\)

\(25B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1980}\right)-\left(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{2005}\right)\)

\(25B=\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{1981}+\frac{1}{1982}+...+\frac{1}{2005}\right)\) (2)

Từ \((1); (2)\Rightarrow 1980A=25B\Rightarrow \frac{A}{B}=\frac{25}{1980}=\frac{5}{396}\)

Ta có: \(A=\dfrac{1}{1.1981}+\dfrac{1}{2.1982}+...+\dfrac{1}{25.2005}\)

\(\Rightarrow1980A=\dfrac{1980}{1.1981}+\dfrac{1980}{2.1982}+...+\dfrac{1980}{25.2005}\)

\(=\dfrac{1}{1}-\dfrac{1}{1981}+\dfrac{1}{2}-\dfrac{1}{1982}+...+\dfrac{1}{25}-\dfrac{1}{2005}\)

\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\)

\(\Rightarrow A=\dfrac{1}{1980}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)

Mặt khác: \(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{1980.2005}\)

\(\Rightarrow25B=\dfrac{25}{1.26}+\dfrac{25}{2.27}+...+\dfrac{25}{1980.2005}\)

\(=\dfrac{1}{1}-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{1980}-\dfrac{1}{2005}\)

\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{1980}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{2005}\right)\)

\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\)

\(\Rightarrow B=\dfrac{1}{25}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)

Do đó A:B=\(\dfrac{1}{1980}:\dfrac{1}{25}=\dfrac{5}{396}\)

Vậy A:B=\(\dfrac{5}{396}\)