\(x^2-4xy+5y^2+10x-22y+28\)
\(=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)
\(=\left(2y-x-5\right)^2+\left(y-1\right)^2+2\)
Ta có :
\(\left(2y-x-5\right)^2\ge0\forall x\)
\(\left(y-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(2y-x-5\right)^2+\left(y-1\right)^2+2\ge2\forall x\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(y-1\right)^2=0\\\left(2y-x-5\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=0\\2y-x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-3\end{matrix}\right.\)
Vậy biểu thức đạt GTNN = 2 ⇔ \(\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
