Có : (x-y)+(y-z)+(x+z) = 2011+(-2012)+2013

=>    x-y+y-z+z-x = 2012

=>  2x=2012

=>x=1006

=>y=1006-2011=-1005

=>z=2013-1006=1007

Chuc ban hoc gioi !!!

Đặt x-y=2011 (1)

     y-z=-2012 (2)

     z+x=2013 (3)

Cộng (1), (2),(3) vế theo vế ta được:

2.x=2012 => x=1006

Từ (1) => y= -1005 

Từ (3) => z= 1007

hình bầu dục

Cộng lại ta có :

\(x-y+y-z+z+x=2011-2012+2013=2012\)

\(=>2x=2012\)

\(=>x=1006\)

Thay vào ta có :

+)\(x-y=2011\)                       

\(=>1006-y=2011\)

\(=>y=1006-2011=-1005\)

+)\(z+x=2013\)

\(=>1006+z=2013\)

\(=>z=2013-1006=1007\)

Vậy x;y;z = 1006;-1005;1007

Ta co: x-y+y-z= 2011+ (-2012)

        <=> x-z=-1 

    Ta co: x-z +z+x= -1+2013

          <=> 2x= 2012 

          <=>x = 1006

Khi do: y= 1006 - 2011=-1005

            z= 2013-1006= 1007

Ta có

x-y=2011 (1)

y-z=-2012 (2)

z+x=2013 (3)

(1)+(2)+(3)=x-y+y-z+z+x=2x=2011+2012+2013=6036

x=6036:2=3018

y=3018-2011=1007

z=1007-(-2012)=3019

Vậy x=3018, y=1007, z=3019

\(\text{Có: }x^2+y^2+z^2=xy+yz+xz\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)=2\left(xy+yz+xz\right)\)

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

\(\Leftrightarrow x^2+x^2+y^2+y^2+z^2+z^2-2xy-2yz-2xz=0\)

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

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

\(\text{Vì }\left(x-y\right)^2\ge0;\left(y-z\right)^2\ge0\text{ và }\left(x-z\right)^2\ge0\)

\(\text{Nên để }\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

\(\text{thì }\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(x-z\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}\Leftrightarrow}x=y=z}\)

\(\text{Khi đó: }x^{2011}+y^{2011}+z^{2011}=3^{2012}\)

\(\Leftrightarrow x^{2011}+x^{2011}+x^{2011}=3^{2012}\left(\text{Vì x = y = z}\right)\)

\(\Leftrightarrow3x^{2011}=3^{2012}\)

\(\Leftrightarrow x^{2011}=3^{2011}\)

\(\Leftrightarrow x=3\)

\(\text{Vậy }x=y=z=3\)