Ta có: \(2x^2+2y^2-x-y-2xy+\frac{1}{2}=0\)
\(\Leftrightarrow\left(x^2+y^2-2xy\right)+\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}^2\right)=0\)
Nhận xét \(\left(x-y\right)^2\ge0;\left(x-\frac{1}{2}\right)^2\ge0;\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-\frac{1}{2}\right)^2=0\\\left(y-\frac{1}{2}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-y=0\\x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Leftrightarrow}x=y=\frac{1}{2}}\)
