Có: \(x^2+y^2+z^2=xy+yz+xz\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2xz\)

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

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

\(\Leftrightarrow\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\)\(\Leftrightarrow x=y=z\)

Lại có: \(x^{2015}+y^{2015}+z^{2015}=3^{2016}\)

\(\Leftrightarrow x^{2015}+x^{2015}+x^{2015}=3^{2016}\)

\(\Leftrightarrow3x^{2015}=3^{2016}\)

\(\Leftrightarrow x=3\)

Vậy \(x=y=z=3\)

ai đó làm giúp mình , mình tích cho

nhân 2 vế cho 2

=>2x²+2y²+2z²=2xy+2yz+2zx

=>2x²+2y²+2z²-2xy-2yz-2zx=0

=>(2x²-2xy)+(2y²-2yz)+(2z²-2zx)=0

=>(x-y)²+(y-z)²+(z-x)²=0

mà (x-y)² >= 0 với mọi x,y

(y-z)² >= 0 với mọi y,z

(z-x)² >=0 với mọi z,x

=>(x-y)²+(y-z)²+(z-x)² >= 0

mà theo đề:(x-y)²+(y-z)²+(z-x)²=0

=>(x-y)²=(y-z)²=(z-x)²=0

=>x=y

   y=z

   z=x

hay x=y=z

do đó x²⁰¹⁵+y²⁰¹⁵+z²⁰¹⁵=3²⁰¹⁶

<=>x²⁰¹⁵+x²⁰¹⁵+x²⁰¹⁵=3²⁰¹⁶

<=>3x²⁰¹⁵=3²⁰¹⁶<=>x²⁰¹⁵=3²⁰¹⁶:3=3²⁰¹⁵<=>x=2015

Vậy x=y=z=2015

cau a ban de o hang dang thuc (x-y-z)^2 di

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

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

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

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

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

Vì mũ chẵn luôn lớn hơn hoặc bằng 0

\(\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Rightarrow}}x=y=z\)

\(\Rightarrow x^{2015}+y^{2015}+z^{2015}=x^{2015}+x^{2015}+x^{2015}=3x^{2015}\)

\(\Rightarrow3x^{2015}=3^{2016}\)

\(\Rightarrow x^{2015}=3^{2015}\)

\(\Rightarrow x=3\)

Vậy \(x=y=z=3\)