Tìm x:
2012×|x-2011|+ (x-2011)^2=2013×|2011-x|
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\(2012.\left|x-2011\right|+\left(x-2011\right)^2=2013\left|2011-x\right|\)
\(2012.\left|x-2011\right|+\left|x-2011\right|^2=2013\left|x-2011\right|\)
\(\left|x-2011\right|\left(2012+\left|x-2011\right|\right)=2013\left|x-2011\right|\)
\(\Rightarrow2012+\left|x-2011\right|=2013\)
\(\left|x-2011\right|=1\)
\(\Rightarrow\orbr{\begin{cases}x=2012\\x=-2010\end{cases}}\)
Ta có: \(\frac{x+4}{2010}+\frac{x+3}{2011}=\frac{x+2}{2012}+\frac{x+1}{2013}\)
\(\Rightarrow\left(\frac{x+4}{2010}+1\right)+\left(\frac{x+3}{2011}+1\right)=\left(\frac{x+2}{2012}+1\right)+\left(\frac{x+1}{2013}+1\right)\)
\(\Rightarrow\left(x+2014\right)\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)
\(\Rightarrow x=-2014\)
=>2012|x-2011|-|x-2011|+(x-2011)^2+2013>0
=>2011|x-2011|+(x-2011)^2+2013>0(luôn đúng)
2012 . | x - 2011| + (x-2011)2 = 2013 . | 2011 - x|
|x-2011|.|x-2011| + 2012 . | x - 2011| - 2013 . | 2011- x| =0
|x - 2011|.| x - 2011| + 2012 .| x - 2011| - 2013 | x - 2011| = 0
| x- 2011| .| x -2011| - | x - 2011| = 0
| x - 2011|. { | x - 2011| - 1} = 0
\(\left[{}\begin{matrix}\left|x-2011\right|=0\\\left|x-2011\right|-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2011\\x=2012\\x=2010\end{matrix}\right.\)
Kết luận x \(\in\) { 2010; 2011; 2012}