Do \(y(y+x)\ne0 \) nên \(y\ne0;y\ne-x\)
Đặt \(t=\dfrac{x}{y},t\ne-1\)
Ta có: \(x^2-xy=2y^2 \Rightarrow(\dfrac{x}{y})^2-\dfrac{x}{y}=2\)
\(\Rightarrow t^2-t-2=0 \Leftrightarrow t=2 \ \ vì \ \ t\ne-1\)
\(\Rightarrow A=\dfrac{1007\dfrac{x}{y}-1}{\dfrac{x}{y}+2012}=\dfrac{2013}{2014}\)
cách khác
\(\left\{{}\begin{matrix}x^2-xy=2y^2\left(1\right)\\y\left(x+y\right)\ne0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x^2-y^2\right)-\left(xy+y^2\right)=\left(x-y\right)\left(x+y\right)-y\left(x+y\right)=\left(x+y\right)\left(x-2y\right)=0\)
Từ (2) =>\(x+y\ne0\Rightarrow x-2y=0\Rightarrow x=2y\)
\(A=\dfrac{1007x-y}{x+2012y}=\dfrac{1007.2y-y}{2y+2012y}=\dfrac{\left(1007.2-1\right)y}{\left(2+2013\right)y}=\dfrac{2013y}{2014y}\)
Từ (2)=> \(y\ne0\) \(\Rightarrow A=\dfrac{2013}{2014}\)
