Thay \(2016=xyz\)vào biểu thức ta được
\(A=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}=\frac{xz+z+1}{xz+z+1}=1\)
Vậy \(A=1\)
Vì \(xyz=2016\)
\(\Rightarrow A=\frac{2016x}{xy+2016x+2016}+\frac{y}{yz+y+2016}+\frac{z}{xz+z+1}\)
\(=\frac{xyz.x}{xy+xyz.x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz+1+z}{xz+z+1}=1\)
