=1 nhưng mình ko biết trình bày bạn nào làm biết trình bày thì viết hô mình nha ?

A=2+2+...+2-2013

=2x1006+2013

=4025

C=(-1+3)+(-5+7)+....+(2011-2013)

  = 2+2+2+...+(-2)

  = 1004+(-2)

  = 1002

D= (2-4)+(6-8)+....+(2010-2012)

  =  -2+-2+-2+...1002+...+-2

  = -502+1002

  = 500

G=(1+2-3-4)+(5+6-7-8)+...+(109+110-111-112)+(113+114+115)

  = -4+-4+-4+...+-4+342

  =-112+342

  = 230 

A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 2005 + 2006 - 2007 - 2008 + 2009 + 2010 ( có 2010 số )

A = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + .... + ( 2005 + 2006 - 2007 - 2008 ) + ( 2009 + 2010 ) 

A = ( - 4 ) + ( - 4 ) + ... + ( - 4 ) + 4019 ( có 503 số )

A = ( - 4 ) . 502 + 4019

A = - 2008 + 4019

A = 2011.

CHÚC LÀM BÀI VUI VẺ

a, C=-1+3-5+7-9+...-2009+2011-2013

=(-1+3)+(-5+7)+...+(-2009+2011)-2013

=(2+2+2+...+2)-2013 (503 số 2)

=2.503-2013=1006-2013=-1007

Vậy C=-1007

b, D=2-4+6-8+...+2006-2008+2010-2012

=(2-4)+(6-8)+...+(2006-2008)+(2010-2012) (503 cặp số)

=-2+(-2)+...+(-2)+(-2) (503 số -2)

=-2.503=-1006

Vậy D=-1006

c, G=1+2-3-4+5+6-7-8+...-111-112+113+114+115

=(1+2-3-4)+(5+6-7-8)+...+(109+110-111-112)+(113+114+115)

=-4+(-4)+...+(-4)+342 (28 số -4)

=-4.28+342=-112+342=230

Vậy G=230

=>\(\left(\dfrac{x^2-8}{2008}-1\right)+\left(\dfrac{x^2-7}{2009}-1\right)=\left(\dfrac{x^2-6}{2010}-1\right)+\left(\dfrac{x^2-5}{2011}-1\right)\)

=>x^2-2016=0

=>x^2=2016

=>\(x=\pm\sqrt{2016}\)

\(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)

\(\left(\frac{x+10}{2008}+1\right)+\left(\frac{x+9}{2009}+1\right)=\left(\frac{x+8}{2010}+1\right)+\left(\frac{x+7}{2011}+1\right)\)

\(\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)

\(\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)

\(x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

mà \(\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)\ne0\)

\(\Rightarrow x+2018=0\)

\(\Rightarrow x=-2018\)

Vậy,.............

Ta có: \(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)

\(\Rightarrow\frac{x+10}{2008}+1+\frac{x+9}{2009}+1=\frac{x+8}{2010}+1+\frac{x+7}{2011}+1\)

\(\Rightarrow\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)

\(\Rightarrow\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)

\(\Rightarrow x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

Do \(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\ne0\)

\(\Rightarrow x+2018=0\)

\(\Rightarrow x=-2018\)

Vậy \(x=-2018\)

\(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}.\)

\(\Rightarrow\frac{x+10}{2008}+\frac{x+9}{2009}-\frac{x+8}{2010}-\frac{x+7}{2011}=0\)

\(\frac{x+10}{2008}+1+\frac{x+9}{2009}+1-\frac{x+8}{2010}-1-\frac{x+7}{2011}-1=0\)

\(\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)

\(\left(x+2018\right).\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

mà \(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\ne0\)

=> x + 2018 = 0

x = -2018