2012=|x-2010|+|x-2008|
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Ta có: x=2011 \(\Rightarrow\)x+1=2012
\(\Rightarrow A=x^{2011}-\left(x+1\right).x^{2010}\)\(+\left(x+1\right)x^{2009}\)\(-\left(x+1\right)x^{2008}+...\)\(-\left(x+1\right)x^2+\left(x+1\right)x-1\)
=\(x^{2011}\)\(-x^{2011}-x^{2010}+x^{2010}+x^{2009}-x^{2009}-\)...\(-x^2+x^2+x-1\)
= \(x-1=2011-1=2010\)
=
+) Xét \(x\ge2010\) có:
\(2012=x-2010+x-2008\)
\(\Rightarrow2x-4018=2012\)
\(\Rightarrow2x=6030\)
\(\Rightarrow x=3015\) ( t/m )
+) Xét \(2008\le x< 2010\) ta có:
\(2012=2010-x+x-2008\)
\(\Rightarrow2=2012\) ( loại )
+) Xét \(x< 2008\) có:
\(2012=2010-x+2008-x\)
\(\Rightarrow2012=4018-2x\)
\(\Rightarrow2x=2006\)
\(\Rightarrow x=1003\left(tm\right)\)
Vậy x= 3015 hoặc x = 1003
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
=>x-2014=0
hay x=2014
`Answer:`
\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)
\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x+2014=0\)
\(\Leftrightarrow x=-2014\)