Tìm x,y thuộc Z để 2y^2x+x+y+1=x^2+2y^2+xy 07/09/2021 Bởi Alexandra Tìm x,y thuộc Z để 2y^2x+x+y+1=x^2+2y^2+xy
$2y^2-2y^2x-x+x^2-y+xy=1\Leftrightarrow2y^2\left(1-x\right)-x\left(1-x\right)-y\left(1-x\right)=1\Leftrightarrow\left(1-x\right)\left(2y^2-x-y\right)=1$ $\Leftrightarrow\left\{{}\begin{matrix}1-x=1\\2y^2-x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y^2-y-1=0\end{matrix}\right.\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right. (tmđk) hay \left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.(ktmđk)Hoặc \left\{{}\begin{matrix}1-x=-1\\2y^2-x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y^2-y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right. (tmđk)hay \left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.$ $Vậy x = 0; y = 1 hoặc x = 2; y = 1$ Bình luận
$2y^2-2y^2x-x+x^2-y+xy=1\Leftrightarrow2y^2\left(1-x\right)-x\left(1-x\right)-y\left(1-x\right)=1\Leftrightarrow\left(1-x\right)\left(2y^2-x-y\right)=1$
$\Leftrightarrow\left\{{}\begin{matrix}1-x=1\\2y^2-x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y^2-y-1=0\end{matrix}\right.\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right. (tmđk) hay \left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.(ktmđk)Hoặc \left\{{}\begin{matrix}1-x=-1\\2y^2-x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y^2-y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right. (tmđk)hay \left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.$
$Vậy x = 0; y = 1 hoặc x = 2; y = 1$