Đặt A = 4x2 + y2 - 4x - 2y + 3
= (4x2 - 4x + 1) + (y2 - 2y + 1) + 1
= (2x - 1)2 + (y - 1)2 + 1 \(\ge1\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,5\\y=1\end{cases}}\)
Vậy Min A = 1 <=> x = 0,5 ; y = 1