K
Khách

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

19 tháng 11 2017

\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(pt\Leftrightarrow\dfrac{1}{100.99}-\left(\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{99.100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}\right)\)

\(=\dfrac{1}{99.100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{99.100}-\left(1-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{99}-\dfrac{1}{100}-1-\dfrac{1}{99}\)

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

20 tháng 11 2017

\(\Rightarrow=\dfrac{1}{100.99}-\left(\dfrac{1}{99.98}+\dfrac{1}{99.97}+...+\dfrac{1}{2.1}\right)\)

\(\Rightarrow\dfrac{1}{100}-\left(\dfrac{1}{99}-\dfrac{1}{98}+\dfrac{1}{98}-....+\dfrac{1}{2}-1\right)\)

\(\Rightarrow\dfrac{1}{100}-\left(\dfrac{1}{99}-1\right)\)

\(\Rightarrow\dfrac{1}{100}-\dfrac{-98}{99}\)

=......... bn tính nhé

19 tháng 11 2017

Gọi \(A=\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow A=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)

\(\Rightarrow A=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)

\(\Rightarrow A=\frac{1}{9900}-\frac{98}{99}=\frac{1}{9900}-\frac{9800}{9900}\)

\(\Rightarrow A=\frac{-9799}{9900}\)

19 tháng 11 2017

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}-\frac{1}{2.1}=-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)

16 tháng 7 2015

\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}\)

=\(\frac{1}{99}-\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}\right)\)

=\(\frac{1}{99}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)\)

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

=\(\frac{1}{99}-\frac{97}{198}\)

=\(\frac{-95}{198}\)

30 tháng 10 2021

\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-...-\dfrac{1}{2.1}\)

\(=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{98}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)

\(=\dfrac{2}{99}-\dfrac{1}{100}-1=-\dfrac{9799}{9900}\)

30 tháng 10 2021

Em chả hiểu j

9 tháng 8 2016

\(M=\frac{1.2.3.4.5...98.99}{10}\)

\(M=1.2.3.4.5.6.7.8.9.11.12...98.99\)

26 tháng 9 2016

Ta xét riêng tử số:

\(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+......+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{1\times99}+\frac{100}{3\times97}+\frac{100}{5\times95}+......+\frac{100}{49\times51}\)

\(=100\times\left(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{49\times51}\right)\)

Bây giờ xét đến mẫu số:

\(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=\frac{2}{1\times99}+\frac{2}{3\times97}+\frac{2}{5\times95}+......+\frac{2}{49\times51}\)

\(=2\times\left(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{49\times51}\right)\)

Vậy giá trị của biểu thức là: \(\frac{100}{2}=50\)

26 tháng 9 2016

thanks 

26 tháng 3 2018

\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+.....+\left(1+2+3+4+......+100\right)}{\left(1.100+2.99+3.98+.......+99.2+100.1\right).2013}\)

\(=\frac{1.100+2.99+3.98+......+99.2+100.1}{\left(1.100+2.99+3.98+.....+99.2+100.1\right).2013}\)

\(=\frac{1}{2013}\)