A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9

các câu 2;3 còn lại giống câu 1 bạn nhé

bạn thay số vào rồi làm tương tự

A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9.

Ta thấy: 2/2.3 = 2/2 - 2/3 ; 2/3.4 = 2/3 - 2/4 ; 2/4.5 = 2/4 - 2/5

Tổng quát ta có: 2/x(x+1) = 2/x - 2/x + 1 , như vậy thì bài toán trên( bạn chép lại đề)

          = 2/1 - 2/x + 1 = 2008/2009

Ta có: 2/1 - 2/x+1 = 2008/2009

2/x+1 =  2 - 2008/2009

2/x+1= 1/2009

x + 1 = 2009

x = 2009 - 1 = 2008

         tk nha

mình đầu tiên và chi tiết nhất! k nha!

Bài này dễ thôi! Tiểu học cũng làm được vì cô Huyền đã hướng dẫn trong Video rồi mà! 

\(\text{#}HaimeeOkk\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}\)

\(A=1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{2019}-\dfrac{1}{2019}\right)-\dfrac{1}{2020}\)

\(A=1-0-0-0-...-0-\dfrac{1}{2020}\)

\(A=1-\dfrac{1}{2020}\)

\(A=\dfrac{2019}{2020}\)

Vậy \(A=\dfrac{2019}{2020}\)

\(M=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{299\cdot300}\)

\(M=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{299}-\frac{1}{300}\)

\(M=\frac{1}{2}-\frac{1}{300}\)

\(M=\frac{149}{300}\)

M = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ ......+\(\frac{1}{299.300}\)

= \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+......+ \(\frac{1}{299}\)- \(\frac{1}{300}\)

= \(\frac{1}{2}\)- \(\frac{1}{300}\)

= \(\frac{149}{300}\)

Lưu ý: Nếu đúng, bạn kết bạn với mình và chọn đúng cho mình nha. Được không?

\(M=\frac{1}{2.3}+\frac{1}{3.2}+\frac{1}{4.5}+...+\frac{1}{299.300}\)

     \(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{299}-\frac{1}{300}\)

     \(=\frac{1}{2}-\frac{1}{300}\)

     \(=\frac{149}{300}\)

\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}+\frac{1}{2017\cdot2018}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2018}\)

\(=2-\frac{1}{2018}\)

\(=\frac{1009}{2018}-\frac{1}{2018}\)

\(=\frac{1008}{2018}=\)TỰ RÚT GỌN NHA

\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2006.2007}+\frac{1}{2007.2008}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2008}\)

\(=2-\frac{2007}{2008}\)

\(=\frac{2009}{2008}\)

~Học tốt~

Xin lỗi, tớ viết sai đề

Làm tiếp khúc sai:

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

\(=\frac{4016}{2008}-\frac{1}{2008}\)

\(=\frac{4015}{2008}\)

~Học tốt~

\(=\frac{1}{2}-\frac{1}{2000}=\frac{999}{2000}\)

Dạng tổng quát :

\(\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{n\left(n+1\right)}=\frac{1}{2}-\frac{1}{n+1}=\frac{n+1-2}{2\left(n+1\right)}=\frac{n-1}{2\left(n+1\right)}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{1999}-\frac{1}{2000}\)

\(=\frac{1}{2}-\frac{1}{2000}\)

\(=\frac{499}{2000}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}\)

                                                                       \(=\frac{1}{2}-\frac{1}{2000}\)

                                                                      \(=\frac{999}{2000}\)

Study well ! >_<

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{10-1}{10}=\dfrac{9}{10}\)

1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=2-1/1.2+3-2/2.3+4-3/3.4+...+10-9/9.10

=1-1/2+1/2-1/3+1/3-1/4+....+1/9-1/10

=1-1/10

=9/10

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)