Đặt (x-2011)^x ra ngoài . Làm tương tự như bài lúc nãy nhé. 

Ta cÓ : ( x - 2011) ^{x+ 1} - ( x - 2011)^{x + 2011}

^{=) x - 2011= 0 =) x = 2011}

x - 2011 = 0 =) x = 2011

chắc là 2011 đó mìh ko chắc nữa

\(\Leftrightarrow2011.\left|x-2011\right|+\left(x-2011\right)^2=2013.\left|x-2011\right|\)

\(\Leftrightarrow\left(x-2011\right)^2=2.\left|x-2011\right|\Rightarrow\orbr{\begin{cases}x-2011=0\\2=\left|x-2011\right|\end{cases}}\Rightarrow\orbr{\begin{cases}x=2013\\x=2009\end{cases}}\text{hoặc }x=2011\)

Vậy ....

\(2011-\left|x-2011\right|=x\)

\(\Rightarrow\left|x-2011\right|=2011-x\)

\(\Rightarrow x-2011=\pm\left(2011-x\right)\)

+) \(x-2011=2011-x\)

\(\Rightarrow2x=2011+2011\)

\(\Rightarrow x=2011\) ( chọn )

+) \(x-2011=-\left(2011-x\right)\)

\(\Rightarrow x-2011=-2011+x\)

\(\Rightarrow0=0\) ( vô lí )

Vậy \(x=2011\)

$2011 - |x-2011| = x$

$\iff |x-2011| = 2011 - x$

$\iff x -2011 \leqslant 0$

$\iff x \leqslant 2011$

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 \(x-2011=0\)

\(x=2011\)

X x 2011-X=2011x2009+2011

Xx(2011-1)=2011x(2009+1)

Xx2010    = 2011x2010

=> X=2011

x=2011

x * 2011 - x = 2011 x 2009 + 2011

x * ( 2011 - 1 ) = 2011 x ( 2009 + 1 )

x * 2010 = 2011 x 2010

x = 2011