Ta có: \(\frac{\left(x-2y\right)^2}{xy}=\frac83\)

=>\(3\left(x-2y\right)^2=8xy\)

=>\(3x^2-12xy+12y^2-8xy=0\)

=>\(3x^2-20xy+12y^2=0\)

=>\(3x^2-18xy-2xy+12y^2=0\)

=>3x(x-6y)-2y(x-6y)=0

=>(x-6y)(3x-2y)=0

TH1: 3x-2y=0

=>3x=2y

=>\(x=\frac23y\)

=>x<y

=>Loại

TH2: x-6y=0

=>x=6y

\(\frac{x}{y}=\frac{6y}{y}=6\)

\(1+x+y=\sqrt{x}+\sqrt{xy}+\sqrt{y}\)

\(\Leftrightarrow2\left(1+x+y\right)=2\left(\sqrt{x}+\sqrt{xy}+\sqrt{y}\right)\) 

\(\Leftrightarrow2+2x+2y=2\sqrt{x}+2\sqrt{xy}+2\sqrt{y}\)

\(\Leftrightarrow2x+2y+2-2\sqrt{x}-2\sqrt{xy}-2\sqrt{y}=0\)

\(\Leftrightarrow\left(x-2\sqrt{xy}+y\right)+\left(x-2\sqrt{x}+1\right)+\left(y-2\sqrt{y}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2=0\)  

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=\sqrt{y}\\\sqrt{x}=1\\\sqrt{y}=1\end{cases}}\)

\(\Leftrightarrow x=y=1\)

\(\Rightarrow S=x^{2013}+y^{2013}=1+1=2\)

Có: \(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=\sqrt{2019}\)

\(\Leftrightarrow\left[xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right]^2=2019\)

\(\Leftrightarrow x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)

\(\Leftrightarrow x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)

\(\Leftrightarrow y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)

\(\Leftrightarrow\left[y\left(1+x^2\right)+x\left(1+y^2\right)\right]^2=2018\)

\(\Leftrightarrow y\left(1+x^2\right)+x\left(1+y^2\right)=\sqrt{2018}\)

hay \(A=\sqrt{2018}\)

\(\frac{1}{2}=\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{2}{xy}\)

\(\Leftrightarrow xy\ge4\)

\(\Rightarrow A=xy+2017\ge4+2017=2021\)