Question

Tìm \(x,y\in Z\) : \(2y^2x+x+y+1=x^2+2y^2+xy\)

Answer 1

Làm lại:

\(2\left(x-1\right)y^2-\left(x-1\right)y=x^2-x-1=x\left(x-1\right)-1\)

với x=1 vô nghiệm

Chia hai vế cho (x-1) khác 0

\(2y^2-y=x-\dfrac{1}{x-1}\)

VP Nguyên x.y, nguyên \(\Rightarrow\dfrac{1}{x-1}\in Z\)

\(\Rightarrow x-1=U\left(1\right)=\left\{-1,1\right\}\Rightarrow x=\left\{0,2\right\}\)

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

Kết luận có các cặp nghiệm: (x,y)=(0,1);(2,1)

Answer 2

\(\left(2y^2x-2y^2\right)+\left(x-xy\right)+\left(1-x^2\right)=0\)

\(\Leftrightarrow2y^2\left(x-1\right)-y\left(x-1\right)-\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2y^2-y-x-1\right)=0\)

\(\left[\begin{matrix}x=1\\2y^2-y-x-1=0\end{matrix}\right.\) ok. {hết thời gian rồi}