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7 tháng 6 2019

Hình như đề bài phải là : Tính tổng : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)

Nếu thế giải như sau : \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}.\)Vậy tổng đó là 2010/2011.

7 tháng 6 2019

Ta có :\(\frac{1}{1}:2+\frac{1}{2}:3+...+\frac{1}{2010}:2011\)

\(\frac{1}{1}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{2010}\times\frac{1}{2011}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(1-\frac{1}{2011}\)

\(\frac{2010}{2011}\)

5 tháng 6 2018

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

\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2009}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2010}\right)\)

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

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}\right)\)

\(\Rightarrow A=\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}\)

\(\Rightarrow\frac{A}{\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}}=1\)

5 tháng 6 2018

Bạn Phạm Tuấn Đạt làm đúng rồi 

Dấu \(.\)là dấu nhân 

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

       \(B=\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}\)

Ta có : 

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

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

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

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}+\frac{1}{2010}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{2010}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}\right)\)

\(\Rightarrow A=\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}\)

\(\Rightarrow A=B\)

Nên : 

\(\frac{A}{B}=\frac{A}{A}=1\)

Vậy giá trị của biểu thức trên là \(1\)

13 tháng 5 2018

Ta có : 

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

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

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}+\frac{1}{2010}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{2010}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1005}\right)\)

\(=\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2010}\)

28 tháng 5 2018

Sửa 1/2+1/3-1/4+...+1/2009-1/2010 thành 1-1/2+1/3-1/4+..+1/2009-1/2010;1/2006+1/2007+..+1/2010 thành 1/1006+1/1007+...+1/2010

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

\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2009}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2010}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2010}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1005}\right)\)

\(\Rightarrow A=\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2010}\)

Ta thấy A = 1/1006+1/2007+...+1/2010

=>A : (1/1006+1/1007+1/1008+...+1/2010) = 1

19 tháng 5 2016

a=1/2.2+1/3.3+1/4.4+...+1/2009.2009+1/2010.2010(có 2009 số hạng)

a=1+1+1+...+1+1(2009 số 1)

a=1.2009=2009

Vậy a>1

18 tháng 3 2023

https://scratch.mit.edu/projects/782275470 

18 tháng 5 2019

\(\frac{1}{1}:2+\frac{1}{2}:3+\frac{1}{3}:4+...+\frac{1}{2009}:2010+\frac{1}{2010}:2011\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{2009}-\frac{1}{2009}\right)+\left(\frac{1}{2010}-\frac{1}{2010}\right)-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}\)

~ Hok tốt ~

18 tháng 5 2019

\(\frac{1}{1}:2+\frac{1}{2}:3+\frac{1}{3}:4+...+\frac{1}{2009}:2010+\frac{1}{2010}:2011\)

\(=\frac{1}{1}:\frac{2}{1}+\frac{1}{2}:\frac{3}{1}+\frac{1}{3}:\frac{4}{1}+...+\frac{1}{2009}:\frac{2010}{1}+\frac{1}{2010}:\frac{2011}{1}\)

\(=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{2009}\cdot\frac{1}{2010}+\frac{1}{2010}\cdot\frac{1}{2011}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2010}+\frac{1}{2010\cdot2011}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}\)

Dấu " . " là dấu nhân nhé