a) \(A=x^2+2y^2+2xy+4x+6y+19\)
\(=\left[\left(x^2+2xy+y^2\right)+2.\left(x+y\right).2+4\right]+\left(y^2+2y+1\right)+14\)
\(=\left[\left(x+y\right)^2+2\left(x+y\right).2+2^2\right]+\left(y+1\right)^2+14\)
\(=\left(x+y+2\right)^2+\left(y+1\right)^2+14\ge14\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y+2=0\\y=-1\end{cases}}\Leftrightarrow x=y=-1\)
b)Đề có gì đó sai sai...
c) Tương tự câu b,em cũng thấy sai sai...HÓng cao nhân giải ạ!
b) \(P=2x^2+y^2+2xy-2y-4\)
\(\Leftrightarrow2P=4x^2+2y^2+4xy-4y-8\)
\(\Leftrightarrow2P=\left(4x^2+4xy+y^2\right)+\left(y^2-4y+4\right)-12\)
\(\Leftrightarrow2P=\left(2x+y\right)^2+\left(y-2\right)^2-12\ge-12\forall x;y\)
Có \(2P\ge-12\Leftrightarrow P\ge-6\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x+y=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)