\(1/1.3+1/3.5+1/5.7+...+1/n.(n+2)<2003/2004\)

Ta có :=2/2.(1/1.3+1/3.5+1/5.7+...+1/n.(n+2)

           =1/2.(2/1.3+2/3.5+2/5.7+...+2/n.(n+2)

           =1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/n-1/n+2)

           =1/2.(1-1/n+2)

           =1/2.(n+2/n+2-1/n+2)

           =1/2.(n+2-1/n+2)

           =1/2.n+1/n+2

           =n+1/(n+2).2

       Vì: n+1/(n+2).2<2003/2004

Suy ra:n+1/(n+2).2=x/2004

Suy ra:(n+2).2=2004

            n+2     =1002

            n         =1000

Vậy n bằng 1000

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

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

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

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Tui hk bít nữa

A = 1 /1.2 + 1/ 2.3 + 1 /3.4 + . . . + 1/ 49.50 + 1/ 50.51

 A = 2 − 1/ 1.2 + 3 − 2 /2.3 + 4 − 3 /3.4 + . . . + 50 − 49 /49.50 + 51 − 50/ 50.51

A = 1 − 1/ 2 + 1/ 2 − 1 /3 + 1 /3 − 1/ 4 + . . . + 1 /50 − 1 /51

A=1-1/51

A=50/51

Cảm ơn bn

B=6/4.7+6/7.10+6/10.13+...+6/73.76

=2.(3/4.7 +3/7.10 +3/10.13 +...+3/73.76 )

=2.(1/4 −1/7 +1/7 −1/10 +1/10 −1/13 +...+1/73 −1/76 )

=2.(1/4-1/76)=2.9/38=9/19

\(S=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)

\(\Rightarrow S=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}\)

\(+..+\frac{\left(1003-1\right)\left(1003+1\right)}{\left(1003.2-1\right)\left(1003.2+1\right)}\)

\(\Rightarrow S=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...\)

\(+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+...+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{1003.2-1}-\frac{1}{1003.2+1}\right)\)

\(\Rightarrow S=1002.\frac{1}{4}-1002.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-...-\frac{1}{1003.2+1}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{2007}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}.\frac{668}{2007}\)

\(\Rightarrow S=\frac{501}{2}-\frac{27889}{223}\)

\(\Rightarrow S=125,4372197\)

\(\)

thx  you

a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)

(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)

(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)

x=\(\frac{1}{3}:\frac{13}{12}\)

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

b)( \(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{9}-\frac{2}{11}_{ }\)):x =\(\frac{2}{3}\)

Giống câu a

=1-(1/3.5+1/3.7+1//7.9+...+1/55.57)

=1-1/2.(2/3.5+2/5.7+2/7.9+...+2/55.57)

=1-1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/55-1/57)

=1-1/2(1/3-1/57)

=1-1/2.18/57

=1-9/57

=48/57

=

1-(1/3.5+1/5.7+1/7.9+....+1/53.55+1/55.57)

=1-1/2.[1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57]

=1-1/2.[1/3-1/57]

=1-1/2.54/171

=1-28/171

=143/171.

Ta có: 1 - 1/3.5 - 1/5.7 - 1/7.9 - ...... - 1/53.55 - 1/55.57

=> 1 - ( 1/3.5 + 1/5.7 + 1/7.9 + ......... + 1/53.55 + 1/55.57 )

=>  1 - ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ....... + 1/53 - 1/55 + 1/55 - 1/57 )

=> 1 - ( 1/3 - 1/57 )

=> 1 - 6/19

=> 13/19

1/3.5+1/5.7+1/7.9+...+1/(2x+1)(2x+3)=5/31

1/2(2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3))=5/31

1/3-1/5+1/5-1/7+1/7-1/9+...+1/2x+1-1/2x+3=5/31:1/2

1/3-1/2x+3=10/31

        1/2x+3=1/3-10/31

        1/2x+3=1/63

suy ra : 2x+3=63 

              2x=63-3 

              2x=60

             x=60:2

             x=30

vay x=30

nhớ **** cho mình nha

ai tk cho mk tròn 30 với

ai tk cho mk 45 chứ ko fải 30 mk nhầm