Ta có :

\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{zx}{z+x}=\frac{xyz}{z\left(x+y\right)}=\frac{xyz}{x\left(y+z\right)}=\frac{xyz}{y\left(x+z\right)}\)

\(\Rightarrow z\left(x+y\right)=x\left(y+z\right)=y\left(z+x\right)\)

Từ \(z\left(x+y\right)=x\left(y+z\right)\Leftrightarrow xz+yz=xy+xz\Leftrightarrow yz=xy\Rightarrow x=z\) (1)

Từ \(x\left(y+z\right)=y\left(x+z\right)\Leftrightarrow xy+xz=xy+yz\Leftrightarrow xz=yz\Rightarrow x=y\) (2)

Từ \(z\left(x+y\right)=y\left(z+x\right)\Leftrightarrow xz+yz=yz+xy\Leftrightarrow xz=xy\Rightarrow z=y\) (3)

Từ (1) ; (2) ; (3) \(\Rightarrow x=y=z\) (đpcm)

\(\frac{2013x}{xy+2013x+2013}+\frac{y}{yz+y+2013}+\frac{z}{xz+z+1}\)

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

\(=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\)

\(=\frac{xz+z+1}{xz+z+1}=1\)

=>đpcm

2013x/xy+2013x+2013 + y/yz+y+2013 + z/xz+z+1

= xyz.x/xy+xyz.x+xyz + y/yz+y+xyz + z/xz+z+1

= xz/1+xz+z + 1/z+1+xz + z/xz+z+1

= xz+1+x/1+xz+x = 1 (đpcm)

Lời giải:

$2023xy+2024yz+4047xz=2023xy+2024y(-x-y)+4047x(-x-y)$

$=-2024y^2-4047x^2-4048xy$

$=-[4047x^2+2024y^2+4048xy]$

$=-[2024(x^2+y^2+2xy)+2023x^2]=-[2024(x+y)^2+2023x^2]$

Vì $2024(x+y)^2+2023x^2\geq 0$ với mọi $x,y$

$\Rightarrow -[2024(x+y)^2+2023x^2]\leq 0$ với mọi $x,y$

Do đó nó không thể nhận giá trị dương.