Question

Giải PT: x² + y² + xy - x - y + 2 = 0

Answer 1

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

\(\Leftrightarrow2x^2+2y^2+2xy-2x-2y+4=0\)

\(\Leftrightarrow x^2+2xy+y^2+x^2+y^2-2x-2y+4=0\)

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

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

Ta thấy \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\forall x,y\\\left(x-1\right)^2\ge0\forall x\\\left(y-1\right)^2\ge0\forall y\\2>0\end{matrix}\right.\Rightarrow\left(x+y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2+2>0\)

Vậy pt vô nghiệm.