\(x^2-3y^2=2xy\)
\(\Leftrightarrow\left(x^2-y^2\right)-\left(2xy+2y^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)
Mà \(x+y\ne0\Rightarrow x-2y=0\Rightarrow x=2y\)
Ta có:\(S=\frac{2x-3y}{4x+5y}=\frac{4y-3y}{8y+5y}=\frac{1}{13}\)