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18 tháng 7 2018

3 câu như nhau cả thôi :v

\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{55\cdot57}\)

\(A=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{55\cdot57}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{55}-\frac{1}{57}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{57}\right)\)

\(A=\frac{1}{2}\cdot\frac{56}{57}\)

\(A=\frac{28}{57}\)

12 tháng 8 2016

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}=\frac{1}{41}\)

=> x + 2 = 41 

=> x = 39

1 tháng 5 2018

\(\frac{1}{1x3}\)\(\frac{1}{3x5}\)+....+\(\frac{1}{9x11}\))                                    x \(y\) = \(\frac{2}{3}\)

\(\frac{2}{1x3}\)\(\frac{2}{3x5}\)+...+\(\frac{2}{9x11}\))                                      x \(y\)\(\frac{4}{3}\)               (nhân 2 vế lên với 2)

(1 - \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)- ...+ \(\frac{1}{9}\)\(\frac{1}{11}\))         x     \(y\)\(\frac{4}{3}\)

( 1 - \(\frac{1}{11}\))                                                                        x    \(y\)=\(\frac{4}{3}\)

\(\frac{10}{11}\)                  x            \(y\)                                                       =\(\frac{4}{3}\)

                                              \(y\)                                                      = \(\frac{4}{3}\)\(\frac{10}{11}\)

                                              \(y\)                                                       = \(\frac{4}{3}\)\(\frac{11}{10}\)

                                               \(y\)                                                       =\(\frac{22}{15}\)

1 tháng 5 2018

kết quả đúng nhưng mình ko hiểu bạn có thể giáng lại ko ?

23 tháng 7 2015

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)

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

\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2}.\frac{6}{7}\)

\(=\frac{3}{7}\)

b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\frac{2010}{2011}\)

\(=\frac{1005}{2011}\)

21 tháng 6 2021

Bạn ơi .là gì thế

 

12 tháng 8 2016

Ta có:

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

                                                          = 1/2.(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2

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

=> 1/2.(1 - 1/x+2) = 20/41

            1 - 1/x+ 2  = 20/41 : 1/2

            1 - 1/x+2   = 40/41

                  1/x+2  = 1/41

=>x + 2 = 41

=>x        = 41 - 2

=>x        = 39

Vậy x = 39

Ủng hộ nha

12 tháng 8 2016

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

=> \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=2.\frac{20}{41}\)

=> \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)

=> \(1-\frac{1}{x+2}=\frac{40}{41}\)

=> \(\frac{1}{x+2}=1-\frac{40}{41}\)

=> \(\frac{1}{x+2}=\frac{1}{41}\)

=> \(x+2=41\)

=> \(x=41-2=39\)

12 tháng 8 2016

Đề sai rồi bạn ơi

Phải là 1/x.(x+2)

12 tháng 8 2016

Gọi tổng trên là A

1/2A= 2/1.3+1/3.5+...+1/x.(x+2)

1/2A= 1-1/x.(x+2)

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

29 tháng 6 2015

Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{50}{101}\)

suy ra: \(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{50}{101}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{x+2}\right)=\frac{50}{101}\)

\(\frac{1}{1}-\frac{1}{x+2}=\frac{50}{101}:\frac{1}{2}=\frac{100}{101}\)

\(\frac{1}{x+2}=1-\frac{100}{101}=\frac{1}{101}\)

suy ra: \(x+2=101\)

suy ra: \(101-2=99\)

7 tháng 6 2016

a) 1/5.6 + 1/6.7 + 1/7.8 + ... + 1/24.25

= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/24 - 1/25

= 1/5 - 1/25

= 4/25

b) 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 -1/101

= 1 - 1/101

= 100/101

c) 3/1.4 + 3/4.7 + ... + 3/2002.2005

= 1 - 1/4 + 1/4 - 1/7 + ... + 1/2002 - 1/2005

= 1 - 1/2005

= 2004/2005

d) 5/2.7 + 5/7.12 + ... + 5/1997.2002

= 1/2 - 1/7 + 1/7 - 1/12 + ... + 1/1997 - 1/2002

= 1/2 - 1/2002

= 500/1001

7 tháng 6 2016

a,A =  \(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{24\times25}\)

A\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

A\(=\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

b, B=\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)

B= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

B=\(1-\frac{1}{101}=\frac{100}{101}\)

c, \(C=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{2002\times2005}\)

C= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)

C= \(1-\frac{1}{2005}=\frac{2004}{2005}\)

d, D= \(\frac{5}{2\times7}+\frac{5}{7\times12}+...+\frac{5}{1997\times2002}\)

D= \(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{1997}-\frac{1}{2002}\)

D= \(\frac{1}{2}-\frac{1}{2002}=\frac{1001}{2002}-\frac{1}{2002}=\frac{1000}{2002}=\frac{500}{1001}\)

26 tháng 8 2019

Tính :

a) \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(=7.\frac{3}{35}\)

\(=\frac{3}{5}\)

c) \(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\frac{2}{75}\)

\(=\frac{1}{75}\)

26 tháng 8 2019

thanks

10 tháng 6 2023

(a+\(\dfrac{1}{1.3}\))+(a+\(\dfrac{1}{3.5}\))+(a+\(\dfrac{1}{5.7}\))+..+(a+\(\dfrac{1}{23.25}\))=11.a+(\(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))

(a+a+..+a)+(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)) = 11.a+ \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))

Đặt A =(a+a+..+a) + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)

Xét dãy số 1; 3; 5;...;25 Dãy số trên là dãy số cách đều với khoảng cách là: 3-1 = 2

Dãy số trên có số số hạng là: (25 - 1): 2 + 1  = 13

Vậy A = a\(\times\)13 + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)

A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{23.25}\))

A = a \(\times\) 13 + \(\dfrac{1}{2}\times\)\(\dfrac{1}{1}-\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)\(\dfrac{1}{7}\)+...+\(\dfrac{1}{23}\) - \(\dfrac{1}{25}\))

A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\) \(\dfrac{24}{25}\)

A = a\(\times\)13 + \(\dfrac{12}{25}\) (1)

Đặt B =    \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\)\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\)

B\(\times\)3 =1 + \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)

B\(\times\)3 - B = 1 - \(\dfrac{1}{243}\) = \(\dfrac{242}{243}\)

2B = \(\dfrac{242}{243}\)

B = \(\dfrac{242}{243}\): 2

B = \(\dfrac{121}{243}\)

11a + B = 11a + \(\dfrac{121}{243}\) (2)

Từ (1) và(2) ta có:

a\(\times\)13  + \(\dfrac{12}{25}\) = 11\(\times\) a + \(\dfrac{121}{143}\)

\(\times\) 13 + \(\dfrac{12}{25}\) - 11 \(\times\)a = \(\dfrac{121}{143}\) 

\(a\times\)(13 - 11) + \(\dfrac{12}{25}\) = \(\dfrac{121}{143}\)

\(\times\) 2 + \(\dfrac{12}{25}\) = \(\dfrac{121}{243}\)

\(\times\) 2 = \(\dfrac{121}{243}\) - \(\dfrac{12}{25}\)

\(\times\) 2 = \(\dfrac{109}{6075}\)

a = \(\dfrac{109}{6075}\): 2

a = \(\dfrac{109}{12150}\)