\(2x^2-4x+4xy+4y^2+4=0\)
\(\Rightarrow x^2+x^2-4x+4xy+4y^2+4=0\)
\(\Rightarrow\left(x^2-4x+4\right)+\left(x^2+4xy+4y^2\right)=0\)
\(\Rightarrow\left(x^2-2.x.2+2^2\right)+\left[x^2+2.x.2y+\left(2y\right)^2\right]\)
\(\Rightarrow\left(x-2\right)^2+\left(x+2y\right)^2=0\)
Ta có :
\(\left(x-2\right)^2\ge0\) \(\forall x\)
\(\left(x+2y\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-2\right)^2+\left(x+2y\right)^2\ge0\)
Dấu = xảy ra khi
\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(x+2y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
