Question

Cho x,y là hai số khác nhau thoả : x²+y =y²+x.Tính \(\dfrac{x^2-y^2+2018xy}{xy-2018}\)

Answer 1

\(x^2+y=y^2+x\\ \Rightarrow x^2+y-y^2-x=0\\ \Leftrightarrow\left(x^2-y^2\right)+\left(y-x\right)=0\\ \Rightarrow\left(x-y\right)\left(x+y\right)-\left(x-y\right)=0\\ \Rightarrow\left(x-y\right)\left(x+y-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-y=0\\x+y-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{x^2-y^2+2018xy}{xy-2018}=\dfrac{2018x^2}{x^2-2018}\)