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Sửa đề : \(\frac{x+3}{2010}+\frac{x+2}{2011}-\frac{x+1}{2012}=\frac{x-1}{2014}\)

\(\Leftrightarrow\frac{x+3}{2010}+1+\frac{x+2}{2011}+1-\frac{x+1}{2012}+1=\frac{x-1}{2014}+1\)

\(\Leftrightarrow\frac{x+3+2010}{2010}+\frac{x+2+2011}{2011}-\frac{x+1+2012}{2012}=\frac{x-1+2014}{2014}\)

\(\Leftrightarrow\frac{x+2013}{2010}+\frac{x+2013}{2011}-\frac{x+2013}{2012}=\frac{x+2013}{1014}\)

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

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

\(\Rightarrow x+2013=0\)( Vì (  \(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2014}\)\(\ne0\))

\(\Rightarrow x=-2013\)

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 \(x-2011=0\)

\(x=2011\)

19 tháng 7 2019

x+1/2009 + x+ 1/2010 + x + 1/2011 = x+1/2012 + x + 1/2013 + x+1/2014

= x+1/2009 + x+1/2010 + x+1/2011 - x+1/2012 - x+1/2013 - x+1/2014 = 0

= (x+1) . ( 1/2009 + 1/2010 + 1/2011 - 1/2012 - 1/2013 - 1/2014) = 0

=x+ = 0 ( Vì 1/2009 + 1/2010 + 1/2011 - 1/2012 - 1/2013 - 1/2014 ≠ 0 )

x=-1

Vậy x=-1

19 tháng 7 2019

Bạn sai rồi

13 tháng 12 2018

\(\Rightarrow\frac{x}{2010}+\frac{x+1}{2011}+\frac{x+2}{2012}+\frac{x+3}{2013}+\frac{x+4}{2014}-5=0\)

\(\left(\frac{x}{2010}-1\right)+\left(\frac{x+1}{2011}-1\right)+\left(\frac{x+2}{2012}-1\right)\)\(+\left(\frac{x+3}{2013}-1\right)+\left(\frac{x+4}{2014}-1\right)=0\)

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

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

mà \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)

Vậy x=-2010

20 tháng 3 2019

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

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

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

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

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

\(\Leftrightarrow x+2014=0\)

\(\Leftrightarrow x=-2014\)

V...

(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 

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!!!!