K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

4 tháng 3 2020

a) \(I=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}\)

\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)

\(I=1-\frac{1}{2010}=\frac{2009}{2010}\)

b) \(K=\frac{4}{2\cdot4}+\frac{4}{2\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\)

\(\frac{1}{2}K=\frac{1}{2}\left(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\right)\)

\(\frac{1}{2}K=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\)

\(\frac{1}{2}K=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2008}-\frac{2}{2010}\)

\(\frac{1}{2}K=1-\frac{1}{2010}=\frac{2009}{2010}\)

\(K=\frac{2009}{2010}:\frac{1}{2}=\frac{2009}{1005}\)

10 tháng 5 2016

\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(K=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(K=2\times\frac{502}{1005}\)

\(K=\frac{1004}{1005}\)

\(F=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\)

\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)

\(3F=\frac{1}{3}-\frac{1}{33}\)

\(F=\frac{10}{33}:3\)

\(F=\frac{10}{99}\)

\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(I=1-\frac{1}{2010}\)

\(I=\frac{2009}{2010}\)

14 tháng 8 2016

dễ mà bạn làm từ câu a nếu ra thì các câu khác cũng dễ thôi

14 tháng 8 2016

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(A=1-\frac{1}{2010}\)

\(A=\frac{2009}{2010}\)

21 tháng 8 2016

\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)

\(\Rightarrow A=4\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{2008.2010}\right)\)

\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\right]\)

\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2010}\right)\right]\Rightarrow A=4\left(\frac{1}{2}.\frac{502}{1005}\right)\Rightarrow A=4.\frac{251}{1005}\Rightarrow A=\frac{1004}{1005}\)

21 tháng 8 2016

\(B=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)

\(\Rightarrow B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+....+\frac{1}{30.33}\)

\(\Rightarrow B=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+.....+\frac{1}{30}-\frac{1}{33}\right)\)

\(\Rightarrow B=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\Rightarrow B=\frac{1}{3}.\frac{10}{33}\Rightarrow B=\frac{10}{99}\)

7 tháng 4 2017

Lâm đi là: 35 phút +2 giờ 20phút =2 giờ 55 phút

7 tháng 4 2017

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

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

\(A=\frac{2016}{2017}\)

23 tháng 4 2017

a,\(\left(3-2\frac{1}{3}+\frac{1}{4}\right):\left(4-5\frac{1}{6}+2\frac{1}{4}\right)\) =\(\left(3-\frac{7}{3}+\frac{1}{4}\right):\left(4-\frac{31}{6}+\frac{9}{4}\right)\)                                                                                                                                    =\(\left(3-\frac{31}{12}\right):\left(4-\frac{1}{3}\right)\)                                                                                                                                                        =\(\frac{5}{12}:\frac{11}{3}\)                                                                                                                                                                              =\(\frac{5}{44}\)                                                                                                      b, F=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.......+\frac{4}{2008.2010}\)                                                                                                                 =\(2.\left(1-\frac{2}{2010}\right)\)                                                                                                                                                                 =\(2.\frac{1004}{1005}\)                                                                                                                                                                             =\(\frac{2008}{1005}\)                                                                                                                                                  

3 tháng 4 2019

a) \(\left(3-2\frac{1}{3}+\frac{1}{4}\right):\left(4-5\frac{1}{6}+2\frac{1}{4}\right)\)

\(=\left(3-\frac{7}{3}+\frac{1}{4}\right):\left(4-\frac{31}{6}+\frac{9}{4}\right)\)

\(=\frac{3-\frac{7}{2}+\frac{1}{3}}{\frac{13}{12}}\)

\(=\frac{11}{13}\)

27 tháng 3 2015

K = 4/2 - 4/4 + 4/4 - 4/6 + ....... + 4/2008 - 4/2010

K = 4/2 - 4/2010

K = 4016/2010 = 1/1003/1005

27 tháng 3 2015

\(\frac{1004}{1005}\)

6 tháng 6 2015

\(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(C=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)\) \(;C=\frac{1}{2}.\frac{502}{1005}=\frac{251}{1005}\)

6 tháng 6 2015

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{1004.1005}\)

=\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}\right)\)

=\(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)

=\(2\left(1-\frac{1}{1005}\right)\)

=\(2.\frac{1004}{1005}\)

=\(\frac{2008}{1005}\)

P/s: Không biết đúng không nữa, làm đại ^.^

4 tháng 12 2017

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2106}\)

\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)

\(A=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)

\(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2014.2016}=\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1007.1008}\right)\)

=> \(B=\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{1008}\right)=\frac{1}{4}.\frac{1007}{1008}\)

=> \(B=\frac{1007}{4032}\)

26 tháng 1 2017

a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{2009.2010}\)

A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)

A=1-\(\frac{1}{2010}\)=\(\frac{2009}{2010}\)

c)C=\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+......+\frac{1}{2006.2008}\)

C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2006}-\frac{1}{2008}\))

C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{2008}\))

C=\(\frac{1}{2}\).\(\frac{1003}{2008}\)=\(\frac{1003}{4016}\)

Câu b mình chưa nghĩ ra

Chúc bạn học tốt!

26 tháng 1 2017

a) A = \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + ...+ \(\frac{1}{2009.2000}\)

= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ... + \(\frac{1}{2009}\) - \(\frac{1}{2000}\)

= 1 - \(\frac{1}{2000}\) = \(\frac{1999}{2000}\)

b) B = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + \(\frac{1}{3.4.5}\) + ... + \(\frac{1}{1998.1999.2000}\)

= \(\frac{1}{2}\) ( \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + \(\frac{2}{3.4.5}\) + ... + \(\frac{2}{1998.1999.2000}\))

= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + \(\frac{1}{3.4}\) - \(\frac{1}{4.5}\) + ... + \(\frac{1}{1998.1999}\) - \(\frac{1}{1999.2000}\))

= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{1999.2000}\))

= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{3998000}\))

= \(\frac{1}{4}\) - \(\frac{1}{7996000}\) = ?

c) C = \(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + \(\frac{1}{6.8}\) + ... + \(\frac{1}{2006.2008}\)

= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{4}\)) + \(\frac{1}{2}\)(\(\frac{1}{4}\) - \(\frac{1}{6}\)) + ... + \(\frac{1}{2}\)(\(\frac{1}{2006}\) - \(\frac{1}{2008}\))

= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\) + ... + \(\frac{1}{2006}\) - \(\frac{1}{2008}\))

= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{2008}\))

= \(\frac{1}{2}\) . \(\frac{1003}{2008}\) = \(\frac{1003}{4016}\).