\(A=\frac{2015x}{xy+2015x+2015}+\frac{y}{yz+y+2015}+\frac{z}{xz+z+1}\)

Thay 2015=xyz vào A, ta được

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

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

\(=\frac{x^2yz+xy+xyz}{xy\left(xz+z+1\right)}=\frac{xy\left(xz+1+z\right)}{xy\left(xz+z+1\right)}=1\)

(x+y+z)(xy+yz+zx)=xyz

x²y+xyz+zx²+xy²+y²z+xyz+xyz+yz²+z²x=xyz

(x²y+xy²)+(xyz+zx²)+(y²z+xyz)+(yz²+z²x)+xyz=xyz

xy(x+y)+zx(y+x)+yz(y+x)+z²(y+x)+xyz=xyz

(x+y)(xy+xz+yz+z²)+xyz=xyz

(x+y)[(xy+xz)+(yz+z²)]+xyz=xyz

(x+y)[x(y+z)+z(y+z)]+xyz=xyz

(x+y)(x+z)(y+z)+xyz=xyz

(x+y)(x+z)(y+z)=xyz-xyz

(x+y)(x+z)(y+z)=0

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

Với x=-z

=>VT= x²⁰¹⁵+y²⁰¹⁵+z²⁰¹⁵=(-z)²⁰¹⁵+z²⁰¹⁵+y²⁰¹⁵=y²⁰¹⁵

VP=(x+y+z)²⁰¹⁵=(-z+y+z)²⁰¹⁵=y²⁰¹⁵

Vậy x²⁰¹⁵+y²⁰¹⁵+z²⁰¹⁵=(x+y+z)²⁰¹⁵ với (x+y+z)(xy+yz+zx)=xyz

Giup mình với nka^^

giúp củ cải

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 \(x=y=z=3\)