Question

[Thanh Quốc]

thấy ko

Answer 1

[Trần Kiều Anh]

ko thay

Answer 2

[Lê Văn Hải]

Ta có: A\(=\)\(2x^2+y^2-2xy+4x+2y+5\)

\(=\left(x^2-2xy+y^2\right)-2\left(x-y\right)+1+\left(x^2+6x+9\right)-5\)

\(=\) \(\left(x-y\right)^2-2\left(x-y\right)+1+\left(x+3\right)^2-5\)

\(=\left(x-y-1\right)^2+\left(x+3\right)^2-5\)

Ta có: \(\left\{\begin{matrix}\left(x-y-1\right)^2\ge0\\\left(x+3\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-y-1\right)^2+\left(x+3\right)^2\ge0\)

\(\Rightarrow\left(x-y-1\right)^2+\left(x+3\right)^2-5\ge-5\)

Hay A \(\ge-5\)

\(\Rightarrow Amin=-5\)

\(\Leftrightarrow\left\{\begin{matrix}x-y-1=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)