Question

giải phương trình nghiệm nguyên

\(2x^2+5y^2-4xy-8y-4x+14=0\)

Answer 1

\(2x^2+5y^2-4xy-8y-4x+14=0\)

\(\Leftrightarrow\left(2x^2+2y^2-4xy\right)+3y^2-8y-4x+14=0\)

\(\Leftrightarrow2\left(x^2+y^2-2xy\right)-4\left(x-y\right)-12y+3y^2+14=0\)

\(\Leftrightarrow2\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]+3y^2-12y+12\)

\(\Leftrightarrow2\left(x-y-1\right)^2+3\left(y-2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)