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

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a) \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)

=> \(\frac{1}{x+1}=\frac{1}{2009}\)

=> x + 1 = 2009

=> x = 2009 - 1

=> x = 2008

b) \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

=> \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

=> \(\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{\left(3x+2\right)\left(3x+5\right)}\right)=\frac{4}{25}\)

=> \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3x+2}-\frac{1}{3x+5}=\frac{4}{25}:\frac{1}{3}\)

=>  \(\frac{1}{2}-\frac{1}{3x+5}=\frac{12}{25}\)

=> \(\frac{1}{3x+5}=\frac{1}{2}-\frac{12}{45}\)

=> \(\frac{1}{3x+5}=\frac{1}{50}\)

=> 3x + 5 = 50

=> 3x = 50 - 5

=> 3x = 45

=> x = 45 : 3

=> x = 15

Đặt \(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2.}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\Rightarrow\frac{S}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(=\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\frac{1}{x+1}=\frac{1}{2001}\)

\(\Rightarrow\)x+1=2001

x=2000

Vậy x=2000.

ĐKXĐ: \(x\ne0;x\ne-1\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2008}{2010}\)

\(\Leftrightarrow2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2008}{2010}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)(Biết công thức này chứ?)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)

\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2008}{2010}\)

\(\Leftrightarrow\dfrac{x-1}{x+1}=\dfrac{2008}{2010}\Leftrightarrow2010x-2010=2008x+2008\Leftrightarrow x=2009\left(tm\right)\)

Vậy x = 2009

d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)

Vậy: x=-1

đặt A = 1.2. + 2.3 + 3.4 + ... + 49.50

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 49.50.3

3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50

3A = 49.50.51

A = 41650

Thay vào ta được

41650 + 1/2x = 40642

=> 1/2x = 1008

=> x = 2016

1+2.( 1/2-1/3+1/3-1/4+....+1/(x-1)-1/x+1)=3/2

1+2.(1/2-1/x+1)=3/2

1-2/x+1=3/2-1

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