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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 \(x-2011=0\)

\(x=2011\)

2 tháng 4 2018

Ta có : 

\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+\frac{x-4}{2009}+\frac{x-2021}{2}=0\)

\(\Leftrightarrow\)\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+\left(\frac{x-4}{2009}-1\right)+\left(\frac{x-2021}{2}+4\right)=0\)

\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+\frac{x-2013}{2009}+\frac{x-2013}{2}=0\)

\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2}\right)=0\)

Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2}\ne0\)

Nên \(x-2013=0\)

\(\Rightarrow\)\(x=2013\)

Vậy \(x=2013\)

Chúc bạn học tốt ~ 

14 tháng 3 2017

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

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

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

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

ta thấy \(\dfrac{1}{2010}>\dfrac{1}{2011}>\dfrac{1}{2012}>\dfrac{1}{2013}\)

=>\(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}>0\)

để A=0

\(\Leftrightarrow x+2014=0\)

\(\Leftrightarrow\)x=-2014

14 tháng 3 2017

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

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

\(\Rightarrow\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)\(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)

\(\Rightarrow x+2014=0\)

\(\Rightarrow x=-2014\)

19 tháng 12 2016

\(\frac{x+1}{2013}+\frac{x}{2012}+\frac{x-1}{2011}=\frac{x-2}{2010}+\frac{x-3}{2009}+\frac{x-4}{2008}\)

\(\Leftrightarrow\frac{x+1}{2013}-1+\frac{x}{2012}-1+\frac{x-1}{2011}-1=\frac{x-2}{2010}-1+\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)

\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}=\frac{x-2012}{2010}+\frac{x-2012}{2009}+\frac{x-2012}{2008}\)

\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}-\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(\Leftrightarrow x-2012=0\). Do \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

\(\Leftrightarrow x=2012\)

16 tháng 3 2018

\(5-\frac{x}{2010}+4-\frac{x}{2011}+3-\frac{x}{2012}=6-\frac{x}{2009}+1-\frac{x}{1007}.\)

\(\left(5+4+3\right)-x.\frac{1}{2010}-x.\frac{1}{2011}-x\frac{1}{2012}=\left(6+1\right)-x.\frac{1}{2009}-x\frac{1}{1007}\)

\(12-x.\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=7-x.\left(\frac{1}{2009}+\frac{1}{1007}\right)\)

\(-x.\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)+x.\left(\frac{1}{2009}+\frac{1}{1007}\right)=7-12\)

\(x.\left(\frac{-1}{2010}-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2009}+\frac{1}{1007}\right)=-5\)

\(x=\frac{-5}{\frac{-1}{2010}-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2009}+\frac{1}{1007}}\)