K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

9 tháng 4 2018

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(\Rightarrow\frac{B}{A}=\frac{2013\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}=2013\)là số nguyên

9 tháng 4 2018

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+..+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

\(\Rightarrow\frac{B}{A}=\frac{\frac{2013}{51}+\frac{2013}{52}+..+\frac{2013}{100}}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}\)

\(=\frac{2013\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}\)

\(=2013\in Z\)

http://olm.vn/hoi-dap/question/126681.html

Bạn tham khảo nhé

10 tháng 4 2017

Ta có:

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)

\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)(1)

Lại có:

\(B\)\(=\dfrac{2013}{51}+\dfrac{2013}{52}+...+\dfrac{2013}{100}\)

\(=2013\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right)\)(2)

Từ (1),(2)\(\Rightarrow\dfrac{B}{A}=2013\)

\(\Rightarrow\dfrac{B}{A}\) là số nguyên

17 tháng 12 2017

Ta có:

A\(=\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+....+\dfrac{1}{99\cdot100}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...\dfrac{1}{99}-\dfrac{1}{100}\)

=\(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}...\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}...\dfrac{1}{100}\right)\)

=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}...+\dfrac{1}{100}\right)\)

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

=\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

Và:

B=\(\dfrac{2013}{51}+\dfrac{2013}{52}+...+\dfrac{2013}{100}\)

=\(2013\cdot\left(\dfrac{1}{51}+\dfrac{1}{52}+...\dfrac{1}{100}\right)\)

\(\Rightarrow\dfrac{B}{A}=2013\)

Vậy\(\dfrac{B}{A}\)là một số nguyên

12 tháng 7 2015

1/1.2 = 2-1/1.2 = 2-1/1 - 2-1/2 = 1-1/2

TƯƠNG TỰ TA CÓ

A = 1-1/2+1/2-1/3+.....+1/99-1/100

A = 1-1/100

A =99/100