\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{x}-\frac{2}{\left(x+2\right)}=\frac{2015}{2016}\)

\(\Rightarrow2-\frac{2}{x+2}=\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{x+2}=2-\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{x+2}=\frac{2017}{2016}\)

\(\Rightarrow2017.\left(x+2\right)=2.2016\)

\(\Rightarrow2017x+4034=4032\)

\(\Rightarrow2017x=-2\)

\(\Rightarrow x=-\frac{2}{2017}\)

Vậy......

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{x\cdot\left(x+2\right)}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)

\(=1-\frac{1}{x+2}=\frac{2015}{2016}\)

=>\(\frac{1}{x+2}=\frac{1}{2016}\)

=>\(x+2=2016\)

=>\(x=2014\)

Vậy.......

Ta có:

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2015}{2016}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{2015}{2016}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{2015}{2016}=\frac{1}{2016}\)

\(\Rightarrow x+2=1\div\frac{1}{2016}=2016\)

\(\Rightarrow x=2016-2=2014\)

Vậy giá trị của x là 2014.

b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

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

\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)

Phương trình tương đương với: 

\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)

c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)

\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)

\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

1) gộp 4 số ra -4 sau đó nhân vào

2) gộp 4 số ra 0

3) mình bó tay

CHÚC BẠN HỌC GIỎI

TK MÌNH NHÉ

a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)

Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=\dfrac{1}{100}\)

\(\Rightarrow x+1=100\)

\(x=100-1\)

\(x=99\)

câu b thiếu kết quả đúng không bn?

Công thức: \(\dfrac{1}{a\times b}=\) 1/ khoảng cách giữa a và b \(\times\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)

* Bạn làm theo công thức và vẫn dụng câu b nhé.

a) Số số hạng của dãy A là: (2020-5):2+1 = 404 (số)

    Tổng A là: (2020+5)x404:2=409050

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

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

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

c) \(C=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)

         \(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+....+\frac{2}{98\times100}\right)\)

           \(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)

             \(=\frac{1}{2}\times\left(1-\frac{1}{100}\right)=\frac{1}{2}\times\frac{99}{100}=\frac{99}{200}\)

Vậy .....

A = 5 + 10 + 15 + ... + 2015 + 2020

Số số hạng là : 404

A = 409050

\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)

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

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

\(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{1}{2}\cdot\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{1}{2}\cdot\left(\frac{1}{98}-\frac{1}{100}\right)\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)

=(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) chia 2 

=(1-1/9)chia 2

=8/9 chia 2

=4/9

Đặt �=11�3+13�5+15�7+17�9A=1x31​+3x51​+5x71​+7x91​

2�=21�3+23�5+25�7+27�92A=1x32​+3x52​+5x72​+7x92​

2�=11−13+13−15+...+17−192A=11​−31​+31​−51​+...+71​−91​

2�=11−19=892A=11​−91​=98​

�=89.12=49A=98​.21​=94​
 

a, Đặt :

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+..............+\dfrac{1}{19.21}\)

\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+............+\dfrac{2}{19.21}\)

\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..........+\dfrac{1}{19}-\dfrac{1}{21}\)

\(\Leftrightarrow2A=1-\dfrac{1}{21}\)

\(\Leftrightarrow2A=\dfrac{20}{21}\)

\(\Leftrightarrow A=\dfrac{10}{21}\)

b, \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...........+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)

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

\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)

\(\Leftrightarrow2A=1-\dfrac{1}{2n+1}\)

\(\Leftrightarrow2A=\dfrac{2n}{2n+1}\)

\(\Leftrightarrow A=\dfrac{n}{2n+1}\)

Đặt \(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}\)

\(2A=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\)

\(2A=\frac{1}{1}-\frac{1}{9}=\frac{8}{9}\)

\(A=\frac{8}{9}.\frac{1}{2}=\frac{4}{9}\)

A= 1/(1x3) + 1/(3x5)+ 1/(5x7) + 1/(7x9) +  1/(9x11)

A x 2 = 2/(1x3) + 2/(3x5)+ 2/(5x7) + 2/(7x9) +  2/(9x11)

Nhận xét :

2/(1x3) = 1 - 1/3

2/(3x5) = 1/3 - 1/5

2/(5x7) = 1/5 - 1/7

2/(7x9) = 1/7 - 1/9

2/(9x11) = 1/9 - 1/11

A x 2 = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11

A x 2 = 1 - 1/11

A x 2 = 10/11

A = 10/11 : 2 = 5/11

các bạn k mình nha!

A= 1/(1x3) + 1/(3x5)+ 1/(5x7) + 1/(7x9) +  1/(9x11)

A x 2 = 2/(1x3) + 2/(3x5)+ 2/(5x7) + 2/(7x9) +  2/(9x11)

Nhận xét :

2/(1x3) = 1 - 1/3

2/(3x5) = 1/3 - 1/5

2/(5x7) = 1/5 - 1/7

2/(7x9) = 1/7 - 1/9

2/(9x11) = 1/9 - 1/11

A x 2 = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11

A x 2 = 1 - 1/11

A x 2 = 10/11

A = 10/11 : 2 = 5/11

Coi A=1/1x3+1/3x5+1/5x7+1/7x9

=>2A=2x(1/1x3+1/3x5+1/5x7+1/7x9)=2/1x3+2/3x5+2/5x7+2/7x9

=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9

=1-1/9=8/9

=>A=8/9:2=4/9