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28 tháng 4 2017

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

Ta tính các số âm và số dương giống nhau cộng lại có tổng bằng 0

\(\Rightarrow A=\frac{2}{2}-\frac{2}{100}\)

\(A=\frac{100}{100}-\frac{2}{100}=\frac{98}{100}=\frac{49}{50}\)

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28 tháng 4 2017

\(A=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+....+\frac{2}{99\cdot100}\)

\(A:2=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{99\cdot100}\)

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

\(A:2=\frac{1}{2}-\frac{1}{100}\)

\(A:2=\frac{49}{100}\)

       A  = \(\frac{49}{50}\)

5 tháng 5 2017

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)

\(\frac{2}{1}\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

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

\(\frac{2}{1}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(\frac{2}{1}.\frac{49}{100}\)

\(\frac{98}{100}=\frac{49}{50}\)

5 tháng 5 2017

Đặt A = \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)

 A : 2 =  \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

 A : 2 = \(\frac{1}{2}-\frac{1}{100}\)

 A : 2 = \(\frac{49}{100}\)

    A   = \(\frac{49}{50}\)

12 tháng 7 2016

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

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

\(=2.\frac{49}{100}\)

\(=\frac{49}{50}\)

12 tháng 7 2016

= 2.(1/2.3 + 1/3.4 + ... + 1/99.100)

trong ngoac co cong thuc do, tim hieu di la lam dc

2 tháng 2 2020

Đặt tổng trên là A , ta có :

\(\frac{A}{2}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(\frac{A}{2}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)

\(\frac{A}{2}=\left(1-\frac{1}{100}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{98}\right)+\left(\frac{1}{99}-\frac{1}{99}\right)\)\(\frac{A}{2}=\frac{99}{100}\)

\(A=\frac{99}{100}.2\)

\(A=\frac{99}{50}\)

30 tháng 10 2016

\(S=\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{98\times99}+\frac{2}{99\times100}\)

\(S=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)

\(S=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(S=2\times\left(1-\frac{1}{100}\right)\)

\(S=2\times\frac{99}{100}\)

\(S=\frac{99}{50}\)

30 tháng 10 2016

\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{98.99}+\frac{2}{99.100}\)

\(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(S=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}+\frac{1}{100}\right)\)

\(S=2.\left(\frac{1}{1}-\frac{1}{100}\right)\\ S=2.\left(\frac{100}{100}+\frac{-1}{100}\right)\\ S=2.\frac{99}{100}\\ S=\frac{99}{50}\)

10 tháng 6 2015

\(1-\frac{2}{2}.3\) hay là \(1-\frac{2}{2.3}\)

4 tháng 2 2017

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

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

\(=2.\frac{98}{99}\)

\(=\frac{196}{99}=1\frac{97}{99}\)

4 tháng 2 2017

Câu b sai rồi

24 tháng 4 2018

\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{99\cdot100}\)

\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\right)\)

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

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

\(=2\cdot\frac{49}{100}\)

\(=\frac{49}{50}\)

24 tháng 4 2018

=2(\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{99.100}\))

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

=2(\(\frac{1}{2}\)-\(\frac{1}{100}\))

=2.\(\frac{49}{100}\)

=\(\frac{49}{50}\)

12 tháng 5 2017

A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

=\(1-\frac{1}{50}\)

Vì \(1-\frac{1}{50}< 1\)nên A < 1

B = \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

=\(\frac{1}{2}-\frac{1}{100}\)

Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)

12 tháng 5 2017

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(\Rightarrow A< 1\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(B=\frac{1}{2}-\frac{1}{100}\)

\(\Rightarrow B< \frac{1}{2}\)

31 tháng 3 2019

Làm bậy, mà đúng

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

31 tháng 3 2019

\(\frac{1}{1.2}\)\(\frac{1}{2.3}\)\(\frac{1}{3.4}\)\(\frac{1}{4.5}\)+ … + \(\frac{1}{99.100}\)

\(\frac{1}{1}\)\(\frac{1}{2}\)\(\frac{1}{2}\)\(\frac{1}{3}\)\(\frac{1}{3}\)-\(\frac{1}{4}\)\(\frac{1}{4}\)\(\frac{1}{5}\)+ … + \(\frac{1}{99}\)\(\frac{1}{100}\)

\(\frac{1}{1}\)\(\frac{1}{100}\)

\(\frac{99}{100}\)