a) \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{24\cdot25}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{25}\)
\(\Leftrightarrow\frac{4}{25}\)

a/ =1/5-1/6+1/6-1/7+1/7-1/8+...+1/24-1/25=1/5-1/25=4/25

Mấy câu còn lại thì từ từ!

b/ \(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)=2\left(1-\frac{1}{101}\right)=2.\frac{100}{101}=\frac{200}{101}\)

Ai thấy đúng thì tik nha Đa tạ!!!!!!!! Chúc bn năm ms zui zẻ!!!!!!!!!!

Ta có: \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2005.2006}\)

\(\Rightarrow N=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2005}-\frac{1}{2006}\)

          \(=1-\frac{1}{2006}=\frac{2005}{2006}\)

 \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2015.2017}\)

      \(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}\)

        \(=1-\frac{1}{2017}=\frac{2016}{2017}\)

N = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/2005 - 1/2006

   = 1/1 - 1/2006

   = 2006/2006 - 1/2006

   =  2005/2006

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+............+\frac{1}{2009}-\frac{1}{2011}=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)

sai rồi top scorer ạ tử trừ mẫu là 2 mà tử là 1 phải nhân 2 lên tử

Ta có: A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2013.2015}\)

\(\Leftrightarrow2A=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2013.2015}\right)\)

\(\Leftrightarrow2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2013}+\dfrac{1}{2013}-\dfrac{1}{2015}\)

\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{2015}=\dfrac{2012}{6045}\)

\(\Leftrightarrow A=\dfrac{1006}{6045}\)

2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{2013.2015}\)

2A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}+\dfrac{1}{2015}\)

2A=\(\dfrac{1}{1}-\dfrac{1}{2015}\)

2A=\(\dfrac{2014}{2015}\)

 A=\(\dfrac{1007}{2015}\)

                     Khi gặp bài này, bn nên tách 1 phân số ra thành hiệu của 2 phân số.

Giải:

A=1/1.3+1/3.5+1/5.7+...+1/2013.2015

A=1/2.(2/1.3+2/3.5+2/5.7+...+2/2013.2015)

A=1/2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/2013-1/2015)

A=1/2.(1/1-1/2015)

A=1/2.2014/2015

A=1007/2015

Chúc bạn học tốt!

a) pt => 2x-x=-25+5(chuyển vế đổi dấu) =>x=-20

b)pt=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\)=\(\frac{2016}{2017}\)

      =>\(1-\frac{1}{2x+1}=\frac{2016}{2017}\)=>\(\frac{2x}{2x+1}=\frac{2016}{2017}\). Nhân chéo => x=1008

3) Ta có : \(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)

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

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

4)

A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

A = \(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{99}-\frac{1}{101}\right)\)

A = \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

A = \(\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(A=\frac{1}{2}.\frac{100}{101}\)

A = \(\frac{50}{101}\)

2, đặt tên biểu thức trên là A. Ta có :

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

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

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

1) \(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}\)

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

\(=1-\frac{1}{5}\)

\(=\frac{4}{5}\)

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