\(2x^2+2y^2+2z^2=2xy+2yz+2zx\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\) \(\Rightarrow x=y=z\)

\(A=\left(2015-2014\right)\left(2014-2013\right)\left(2013-2012\right)=1\)

x,y,z=0

Đặt \(\frac{x}{2011}=\frac{y}{2012}=\frac{z}{2013}=k\)

\(\Rightarrow\hept{\begin{cases}x=2011k\\y=2012k\\z=2013k\end{cases}}\)

+) Ta có : \(\frac{2012z-2013y}{2011}=\frac{2012.2013k-2013.2012k}{2011}=0\)

\(\frac{2013x-2011z}{2012}=\frac{2013.2011k-2011.2013k}{2012}=0\)

\(\frac{2011y-2012x}{2013}=\frac{2011.2012k-2012.2011k}{2013}=0\)

Do đó : \(\frac{2012z-2013y}{2011}=\frac{2013x-2011z}{2012}=\frac{2011y-2012x}{2013}\left(=0\right)\) ( đpcm )

Đặt \(\frac{x}{2011}=\frac{y}{2012}=\frac{z}{2013}=k\Rightarrow\hept{\begin{cases}x=2011k\\y=2012k\\z=2013k\end{cases}}\)

\(\frac{2012z-2013y}{2011}=\frac{2012\cdot2013k-2013k\cdot2012}{2011}=\frac{0}{2011}=0\)(1)

\(\frac{2013x-2011z}{2012}=\frac{2013\cdot2011k-2011\cdot2013k}{2012}=\frac{0}{2012}=0\)(2)

\(\frac{2011y-2012x}{2013}=\frac{2011\cdot2012k-2012\cdot2011k}{2013}=\frac{0}{2013}=0\)(3)

Từ (1) , (2) và (3) => đpcm

\(\frac{x}{y}=\frac{2}{3}\Rightarrow x=\frac{2y}{3}\)

Thế vào A ta được: \(A=\frac{2014x+2013y}{2014x-2013y}=\frac{2014.\frac{2y}{3}+2013y}{2014.\frac{2y}{3}-2013y}=\frac{y\left(2014.\frac{2}{3}+2013\right)}{y\left(2014.\frac{2}{3}-2013\right)}\)

                             \(A=\frac{\frac{10067}{3}}{\left(-\frac{2011}{3}\right)}=\frac{-10067}{2011}\)

P/s: Không chắc lắm

Vì \(\dfrac{x}{y}=\dfrac{2}{3}->x=2,y=3\)

A =\(\dfrac{2014.2+2014.3}{2014.2-2014.3}=\dfrac{4028+6042}{4028-6042}=\dfrac{10070}{-2014}=-5\)

\(\sqrt{x-2013}+x^3=\sqrt{y-2013}+y^3\)

\(\Leftrightarrow\sqrt{x-2013}-\sqrt{y-2013}+x^3-y^3=0\)

\(\Leftrightarrow\dfrac{x-y}{\sqrt{x-2013}+\sqrt{y-2013}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x-2013}+\sqrt{y-2013}}+\left(x^2+xy+y^2\right)\right)=0\)

\(\Leftrightarrow x=y\)

\(\Rightarrow B=\dfrac{2013x+2014y}{2013y+2014x}=1\)

Đặt \(A=\dfrac{2014x}{xy+2014x+2014}+\dfrac{y}{yz+y+2014}+\dfrac{z}{xz+z+1}\)

\(A=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\)

\(A=\dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}\)

\(A=\dfrac{xz}{xz+z+1}+\dfrac{1}{xz+z+1}+\dfrac{z}{xz+z+1}\)

\(A=\dfrac{xz+z+1}{xz+z+1}=1\)

\(\Rightarrowđpcm\)

Ta có : \(A=\dfrac{2014x}{xy+2014x+2014}+\dfrac{y}{yz+y+2014}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{xyz.x}{xy+xyz.x+xyz}+\dfrac{x.y}{x.yz+xy+xyz.x}+\dfrac{xy.z}{xz.xy+xy.z+xy}\)

\(=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{xy}{xyz+x^2yz+xy}+\dfrac{xyz}{x^2yz+xyz+xy}\)

\(=\dfrac{x^2yz+xyz+xy}{x^2yz+xyz+xy}=1\) (const)

Vậy A không phụ thuộc vào các biến x,y,z