Tag thầy Lâm không :)???

!x-2007!+!x-2010!>=3 đẳng thức khi 2007<=x<=2008 

!x-2007!+!x-2008!+!x-2010!>=3 đẳng thức khi !x-2008!=0 

=> nghiệm duy nhất x=2008 và y=2009

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

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

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

=>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

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

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

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

\(\Rightarrow x+2014=0\)

\(\Rightarrow x=-2014\)

tìm x biết :

a)(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

Vì : 1/2010 + 1/2011 - 1/2012 - 1/2013 khác 0

==> x + 2014 = 0

==> x = - 2014

Vậy x = - 2014.

Chúc pạn hok tốt!!!

\(\Rightarrow\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x=-3\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\ne0\right)\)

\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)

\(\Leftrightarrow x+3=0\)

hay x=-3

\(\left(\frac{x+4}{2007}+1\right)+\left(\frac{x+3}{2008}+1\right)=\left(\frac{x+2}{2009}+1\right)+\left(\frac{x+1}{2010}+1\right)\)

\(\left(\frac{x+2011}{2007}\right)+\left(\frac{x+2011}{2008}\right)=\left(\frac{x+2011}{2009}\right)+\left(\frac{x+2011}{2010}\right)\)
\(\frac{x+2011}{2007}+\frac{x+2011}{2008}-\frac{x+2011}{2009}-\frac{x+2011}{2010}=0\)

\(\left(x+2011\right).\left(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)

Vì \(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\)khác 0 (các số hạng ko bằng nhau)

\(\Leftrightarrow\)\(x+2011=0\)

\(\Rightarrow x=0-2011\)

\(\Rightarrow x=-2011\)

\(\left|x-2007\right|+\left|x-2010\right|+\left|x-2008\right|+\left|y-2009\right|\)

\(\ge\left|x-2007+2010-x\right|+\left|x-2008\right|+\left|y-2009\right|=3+0+0=3\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-2007\right)\left(2010-x\right)\ge0\\\left|x-2008\right|=0\\\left|y-2009\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2008\\y=2009\end{cases}}\)

Vậy x = 2008 và y = 2009

\(\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\)

\(\Rightarrow\left|x-2017\right|+\left|x-2018\right|+\left|2010-x\right|+\left|y-2009\right|=3\)

Ta có :+) \(\left|x-2007\right|+\left|2010-x\right|\ge\left|x-2007+2010-x\right|=3\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-2007\right)\left(2010-x\right)\ge0\Leftrightarrow2007\le x\le2010\)

+) \(\left|x-2008\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow x-2008=0\Leftrightarrow x=2008\)

+)\(\left|y-2009\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow y-2009=0\Leftrightarrow y=2009\)

\(\Rightarrow\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|\ge3\)

\(\Rightarrow\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\)

\(\Leftrightarrow\hept{\begin{cases}2007\le x\le2010\\x=2008\\y=2009\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2008\\y=2009\end{cases}}\)

Vậy................................