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Đặt 2017x-2016=a; 2016x-2015=b
Theo đề, ta có: \(a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow x\in\left\{\dfrac{2016}{2017};\dfrac{2015}{2016};\dfrac{4031}{4033}\right\}\)
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 \(xyz=2016\)
\(\Rightarrow A=\frac{2016x}{xy+2016x+2016}+\frac{y}{yz+y+2016}+\frac{z}{xz+z+1}\)
\(=\frac{xyz.x}{xy+xyz.x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz+1+z}{xz+z+1}=1\)
Ta có: \(\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)\\ =\left(xy+\left(x+y+z\right)z\right)\left(yz+\left(x+y+z\right)x\right)\left(zx+\left(x+y+z\right)y\right)\\ =\left(xy+zx+zy+z^2\right)\left(yz+x^2+xy+xz\right)\left(zx+xỹ+y^2+yz\right)\\ =\left(y+z\right)\left(x+z\right)\left(x+z\right)\left(y+x\right)\left(z+y\right)\left(x+y\right)\\ =\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2\\ \Rightarrow\frac{\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =\frac{\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =1\)
x=2015
=> x+1=2016
=> A=x2016-(x+1).x2015+(x+1).x2014-(x+1).x2013+...+(x+1)x2-(x+1)x+2016
=x2016-x2016-x2015+x2015+x2014-x2014-x2013+...+x3+x2-x2-x+2016
=-x+2016
=-2015+2016
=1
Vậy A=1.