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

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

= \(1-\frac{1}{50}=\frac{49}{50}\)

B = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

= \(2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\right)\)

= \(2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)

= \(\frac{2}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)

= \(\frac{4}{13}\)

C = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

= \(3\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)

= \(3.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)

= \(\frac{3}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\) 

= \(\frac{9}{38}\)

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

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

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

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

Ta có: \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2005.2006}\)

\(\Rightarrow N=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2005}-\frac{1}{2006}\)

          \(=1-\frac{1}{2006}=\frac{2005}{2006}\)

 \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2015.2017}\)

      \(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}\)

        \(=1-\frac{1}{2017}=\frac{2016}{2017}\)

N = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/2005 - 1/2006

   = 1/1 - 1/2006

   = 2006/2006 - 1/2006

   =  2005/2006

A=1/1*3+1/3*5+...+1/2017*2019

2A=2/1*3+2/3*5+...+2/2017*2019

2A=1-1/3+1/3-1/5+..+1/2017-1/2019

2A=1-1/2019

2A=2018/2019

A=(2018/2019):2

A=1009/2019

A=1009/2019

\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2017.2019}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2017.2019}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2019}\right)\)

\(A=\frac{1}{2}.\frac{2018}{2019}\)

\(A=\frac{1009}{2019}\)

A) [124 - (20 - 4x)] = 12 . 20 

[124 - (20 - 4x)] = 240 

(20 - 4x)  = 240 - 124

(20 - 4x) = 116

4x = 116 + 20

4x = 136 

x = 136 : 4 

x = 34

B) (2x - 1) = 1/3 : -4/21

2x - 1 = -7/4

2x = -7/4 + 1 

2x = -3/4

x = -3/4 : 2 

x = -3/8

ta có 

\(C=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{4.3}+..+\frac{100-99}{99.100}\)

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

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

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

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

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

giúp vs

= 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +... + 1/99 - 1/100

= 1 - 99/100

= 1/100.

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

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

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

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

Mọi thứ trong thế giới này chỉ là một trò chơi và chúng ta chỉ là những con tốt...

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