\(4x^2+4y-4xy+5y^2+1=0\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(4y^2+4y+1\right)=0\Leftrightarrow\left(2x-y\right)^2+\left(2y+1\right)^2=0\) Vì \(\left(2x-y\right)^2\ge0\forall x,y\) ; \(\left(2y+1\right)^2\ge0\forall y\) Nên \(\left(2x-y\right)^2+\left(2y+1\right)^2=0\) khi:
\(\left[{}\begin{matrix}\left(2x-y\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-y=0\\2y+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=y\\2y=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=y\\y=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\y=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy \(x=\dfrac{-1}{4};y=\dfrac{-1}{2}\)
