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

b, \(\frac{x+1}{2009}+\frac{x+2}{2009}=\frac{x+10}{2000}+\frac{x+11}{1999}\)

\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)\)

\(\Rightarrow\frac{x+1+2009}{2009}+\frac{x+2+2008}{2008}=\frac{x+10+2000}{2000}+\frac{x+11+1999}{1999}\)

\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2000}+\frac{x+2010}{1999}\)

\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2000}-\frac{x+2010}{1999}=0\)

\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\right)=0\)

Mà \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\ne0\)

=> x + 2010 = 0 => x = -2010

ai la Fc cua lam chan khang kb duoc khong?

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

\(\Leftrightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

Để |x+2007|+|x+2006|=1

Ta có |x+2007|>|x+2006|. Suy ra x+2007=1 và x+2006=0 suy ra x=-2006

Ai k mình k lại gấp 10

x = - 2007

1/x+x+1+x+2+x+3+...+x+2006+2007=2007

------------------------------------------=2007-2007

------------------------------------------=0

x+x+x+...+x+1+2+3+...+2006=0

2007.x+(1+2+...+2006)=0

2007.x+(2006+1).[(2006-1)+1]:2=0

2007.x+2013021=0

2007.x=0-2013021

x=-2013021:2007

x=-1003

2/x+x+1+x+2+...+x+198=401-201-200-199

199.x+(1+2+...+198)=-199

199.x+(1+198).[(198-1)+1]:2=-199

199.x+19701=-199

199.x=-199-19701

x=-19900:199

x=-100

3/x+x+1+x+2+...+x+2008=2010-2010-2009

2009.x+(2008+1).[(2008-1)+1]:2=-2009

2009.x+2017036=-2009

2009.x=-2009-2017036

x=-2019045:2009

x=-1005

\(x-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{2006}-\frac{x}{2007}=\frac{2006}{2007}\)

\(x-\frac{x}{2007}=\frac{2006}{2007}\)

\(\frac{2007x-x}{2007}=\frac{2006}{2007}\)

\(\frac{2006x}{2007}=\frac{2006}{2007}\Rightarrow2006x=2006\)

=>x=1

Các số chẵn từ 2 đến 2000 là (2000-2)/2+1=1000
Các số lẻ từ 1 đến 2001 là (2001-1)/2+1=1001

1000 số chẵn từ 2 đến 2000 có 1000 số lẻ liền sau từ 3 đến 2001, nên tổng các số lẻ từ 3 đến 2001 hơn tổng các số chẵn từ 2 đến 2000 là 1000.
Thêm 1 là số lẻ nữa, suy ra tổng các số lẻ từ 1 đến 2001 hơn tổng các số chẵn là 1001