428758

\(S=\frac{101}{120}+\frac{1}{2.3}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{18.19}+\frac{1}{19.20}\right)\)

\(S=\frac{101}{120}+\frac{1}{6}\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{19-18}{18.19}+\frac{20-19}{19.20}\right)\)

\(S=\frac{101}{120}+\frac{1}{6}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

\(S=\frac{101}{120}+\frac{1}{6}\left(1-\frac{1}{20}\right)=\frac{101}{120}+\frac{19}{120}=\frac{120}{120}=1\)

= 2.(1 / 2.3 + 1 / 3.4 + ..... + 1 / x (x + 1) = 2007/2009

= 2.(1/2 - 1/3 + 1/3 - +.......+ 1/x - 1/x+1) = 2007/2009

= 2.( 1/2 - 1/x+1) = 2007/2009

= 1 - 1/x+1 =2007/2009

= 1/x+1 = 1/2009

=> x + 1 = 2009

=> x = 2008

Ta có: 2/2.3 + 2/3.4 + .... + 2/x.(x+1) = 2007/2009

=> 2.[1/2.3+1/3.4+.....+1/x.(x+1)]=2007/2009

=> 2.(1/2-1/3+1/3-1/4 + .... + 1/x - 1/x+1) = 2007/2009

=> 2.(1/2-1/x+1)=2007/2009

=>1/2 - 1/x+1 = 2007/2009 : 2

=> 1/2 - 1/x+1 = 2007/4018

=> 1/x+1 = 2007/4018 +1/2 

=> 1/x+1 = 

= 2.(1 / 2.3 + 1 / 3.4 + ..... + 1 / x (x + 1) = 2007/2009

= 2.(1/2 - 1/3 + 1/3 - +.......+ 1/x - 1/x+1) = 2007/2009

= 2.( 1/2 - 1/x+1) = 2007/2009

= 1 - 1/x+1 =2007/2009

= 1/x+1 = 1/2009

=> x + 1 = 2009

=> x = 2008

Đặt tổng trên là A

\(5A=\frac{5}{4x9}+\frac{5}{9x14}+\frac{5}{14x19}+...+\frac{5}{44x49}\)

\(5A=\frac{9-4}{4x9}+\frac{14-9}{9x14}+\frac{19-14}{14x19}+...+\frac{49-44}{44x49}\)

\(5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\)

\(5A=\frac{1}{4}-\frac{1}{49}\Rightarrow A=\frac{49-4}{4x5x49}=\frac{45}{4x5x49}=\frac{9}{4x49}\)

\(\frac{1}{2x4}+\frac{1}{4x6}+...+\frac{1}{96x98}+\frac{1}{98x199}=\frac{2}{2x4}+\frac{2}{4x6}+...+\frac{2}{99x100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

A x2 = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{100}\)

A x2 = \(\frac{49}{100}\)

A = \(\frac{49}{200}\)

1/2*(2/2*4+2/4*6+...+2/98*100)=1/2*(1/2-1/4+1/4-1/6+...+1/98-1/100)

                                            =1/2*(1/2-1/100)

                                            =1/2*49/100

                                            =49/200

k nha bạn

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{96.98}+\frac{1}{98.100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\)

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

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

cho mình tròn 1550 nhé bạn

=2/2 4+2/4 6+2/6 8+...+2/96 98+2/98 100

=2/2-2/4+2/4-2/6+2/6-2/8+...+2/96-2/98+2/98-2/100

=2/2-2/100

=99/100

\(A=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\)

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

\(=\frac{50}{100}-\frac{1}{100}\)

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

Vậy: \(A=\frac{49}{100}\)

Ta có:\(2A=2\left(\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{98.100}\right)\)

\(=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{98.100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)

Vậy giá trị của A là \(\frac{49}{200}\)

THANKS NHÉ!