A=y2-2y(x-1)+4x2-10x+14
A=y2-2y(x-1)+(x-1)2-(x-1)2+4x2-10x+14
A=(y-x+1)2-x2+2x-1+4x2-10x+14
A=(y-x+1)2+3x2-8x+13
A=(y-x+1)2+3(x2-2x\(\frac{4}{3}\)+\(\frac{16}{9}\))-\(\frac{16}{3}\)+13
A=(y-x+1)2+3(x-\(\frac{4}{3}\))2+\(\frac{23}{3}\)\(\ge\)\(\frac{23}{3}\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}x-\frac{4}{3}=0\\y-x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=\frac{1}{3}\end{matrix}\right.\)
Vậy GTNN của A là \(\frac{23}{3}\) khi x=\(\frac{4}{3}\), y=\(\frac{1}{3}\)
\(A=4x^2+y^2-10x-2xy-2y+14\)
\(=\left(y^2-2xy-2y\right)+4x^2-10x+14\)
\(=y^2-2y\left(x+1\right)+\left(x+1\right)^2-x^2-2x-1+4x^2-10x+14\)
\(=\left(y-x-1\right)^2+3x^2-12x+13\)
\(=\left(y-x-1\right)^2+3\left(x^2-4x+4\right)+1\)
\(=\left(y-x-1\right)^2+3\left(x-2\right)^2+1\ge1\forall x;y\)
(do...)
\(\Rightarrow MinA=1\Leftrightarrow\)\(\left\{{}\begin{matrix}y-x=1\\x=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)
