Đặt x -2006 = y 

pt <=>  \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)

<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)

<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)

<=> \(49y^2-49y+49=57y^2-57y+19\)

<=> \(8y^2-8y-30=0\)

<=> \(4y^2-4y+15=0\)

Giải tiếp nha 

Đặt vế trái là A ta có:

\(\frac{A}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)

\(\frac{A}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)

\(\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow A=\frac{x-1}{x+1}\)

\(\Rightarrow\frac{x-1}{x+1}=\frac{2007}{2009}\Leftrightarrow x=2003\)
 

​\(\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow...

Đặt vế trái là A ta có:

\frac{A}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}2A​=61​+121​+201​+...+x(x+1)1​=2.31​+3.41​+4.51​+...+x(x+1)1​

\frac{A}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{x+1-x}{x\left(x+1\right)}=\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}2A​=2.33−2​+3.44−3​+4.55−4​+...+x(x+1)x+1−x​=21​−31​+31​−41​+41​−51​+...+x1​−x+11​

\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow A=\frac{x-1}{x+1}2A​=21​−x+11​⇒2A​=2(x+1)x+1−2​⇒A=x+1x−1​

\Rightarrow\frac{x-1}{x+1}=\frac{2007}{2009}\Leftrightarrow x=2003⇒x+1x−1​=20092007​⇔x=2003

ta có: 1/3 + 1/6 + ... + 2/x(x+1) = 2/2.3 + 2/3.4 +.......2/x(x+1) = 2(1/2.3 +1/3.4 +.....+1/x(x+1)) = 2.(1/2-1/3+1/3-1/4+....+1/x-1/(x+1))= 2.(1/2-1/(x+1)) = 1-2/(x+1)

giải 1-2/(x+1) = 2007/2009 ta được x=2008

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\)

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

\(=2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{x+1-x}{x\left(x+1\right)}\right)\)

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

\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)\)

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

Phương trình ban đầu tương đương với: 

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

\(\Leftrightarrow x=2008\). 

http://h.vn/hoi-dap/question/94327.html

mình bó tay

\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\right)=0\)

\(\Leftrightarrow x=2010\)

a) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 2006 ) + 2007 = 2007

\(\Rightarrow\)( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 2006 + 2007 ) = 2007

\(\Rightarrow\)2007x + 2015028 = 2007

\(\Rightarrow\)2007x = 2007 - 2015028 = -2013021

\(\Rightarrow\)x = ( -2013021 ) : 2007 = -1003

Vậy x = -1003

b) 2000 + ( 199 + x ) + ( 198 + x ) + ... + ( x + 1 ) + x = 200

\(\Rightarrow\)( x + x + x + ... + x + x ) + ( 1 + 2 + ... + 198 + 199 + 2000 ) = 200

\(\Rightarrow\)200x + 2001000 = 200

\(\Rightarrow\)200x = 200 - 2001000 = -2000800

\(\Rightarrow\)x = ( -2000800 ) : 200 = -10004

Vậy x = -10004 

a, x + ( x + 1 ) + ( x + 2 ) + ..... + ( x + 2006) + 2007 = 2007

x. 2007 + ( 1 + 2 + ..... + 2006 ) = 2007 - 2007

x. 2007 + 2013021 = 0

x. 2007                 = 0 - 2013021

x.2007                  = - 2013021

    x                       = ( - 2013021 ) : 2007

   x                        = - 1003

a) x + ( x + 1 ) + ( x + 2) + ........+ ( x + 2006 ) +2007 =2007 

(x+x+x+...+x)+(1+2+3+...+2007)                                =2007

2007.x          + [(2007+1).2007 : 2]                              =2007

2007.x         +2015028                                                =2007

2007.x                                                                          =2007-2015028

x                                                                                   =(-2013021) : 2007

x                                                                                   = -1003

Vây   x = -1003