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1 tháng 4 2016

Ta có: \(\frac{x+4}{2010}+\frac{x+3}{2011}=\frac{x+2}{2012}+\frac{x+1}{2013}\) 

\(\Rightarrow\left(\frac{x+4}{2010}+1\right)+\left(\frac{x+3}{2011}+1\right)=\left(\frac{x+2}{2012}+1\right)+\left(\frac{x+1}{2013}+1\right)\)

\(\Rightarrow\left(x+2014\right)\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)

\(\Rightarrow x=-2014\)

(x + 4)/2010 + (x+3)/2011 = (x+2)/2012 + (x+1)/2013 

<=> [(x + 4)/2010 + 1] + [(x+3)/2011 + 1] = [(x+2)/2012 + 1] + [(x+1)/2013 + 1] 

<=> (x + 2014)/2010 + (x + 2014)/2011 = (x + 2014)/2012 + (x + 2014)/2013 

<=> (x + 2014)/2010 + (x + 2014)/2011 - (x + 2014)/2012 - (x + 2014)/2013 = 0 

<=> (x + 2014).(1/2010 + 1/2011 - 1/2012 - 1/2013) = 0 

Ta thấy (1/2010 + 1/2011 - 1/2012 - 1/2013) ≠ 0 

Vậy suy ra x = -2014

6 tháng 4 2017

đúng ko các bạn 

5 tháng 7 2023

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)

\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)

\(\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)

Vì \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\) 

=> \(x+2014=0\) 

                  \(x=0-2014\) 

                  \(x=-2014\)

9 tháng 8 2018

\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}=\frac{x-4}{2010}\)

\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}-\frac{x-4}{2010}=0\)

\(\frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\frac{x-3}{2011}+1-\frac{x-4}{2010}+1=0\)

\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)-\left(\frac{x-3}{2011}-1\right)-\left(\frac{x-4}{2010}-1\right)=0\)

\(\frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}-\frac{x-2014}{2010}=0\)

\(\left(x-2014\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)

\(x-2014=0:\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)\)

\(x-2014=0\)

\(x=2014\)

Nhớ tk cho mình nha =3

A.R.M.Y   FIGHTING!!!!

21 tháng 3 2023

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)

\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)

\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)

`=> (x+2014) (1/2010 + 1/2011-1/2012-1/2013)=0`

`=> x+2014=0` ( vì `1/2010 + 1/2011-1/2012-1/2013≠0 )`

`=>x=-2014`

 

29 tháng 11 2016

Ta có: \(\frac{x-1}{2013}+\frac{x-2}{2012}=\frac{x-3}{2011}+\frac{x-4}{2010}\)

\(\Rightarrow\frac{x-1}{2013}+1+\frac{x-2}{2012}+1=\frac{x-3}{2011}+1+\frac{x-4}{2010}+1\)

\(\Rightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2014}{2011}+\frac{x-2014}{2010}=0\)

\(\Rightarrow\left(x-2014\right).\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2011}\right)=0\)

Vì \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\ne0\)

=> x - 2014 =0

=> x          = 2014

Vậy x = 2014

12 tháng 2 2016

ủng hộ mình lên 280 điểm với các bạn

\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

vì \(\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\Rightarrow x+2004=0\Rightarrow x=-2004\)

vậy x=-2004

17 tháng 9 2018

a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)

\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)

\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)

\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)    (1)

Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)

Nên biểu thức (1) xảy ra khi \(x+2013=0\)

\(x=-2013\)

b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)  (2)

Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)

Nên biểu thức (2) xảy ra khi \(x-2011=0\)

\(x=2011\)