\(A=2x^2+2xy+y^2-2x+2y+1\)
\(A=\left(x^2+2xy+y^2\right)+2\left(x+y\right)+1+x^2-4x+4-4\)
\(A=\left(x+y\right)^2+2\left(x+y\right)+1+\left(x+2\right)^2-4\)
\(A=\left(x+y+1\right)^2+\left(x+2\right)^2-4\ge4\)
Dấu "=" xảy ra khi: x=2, y=-3
ta có 2A=\(4x^2+4xy+2y^2-4x+4y+2=4x^2+y^2+1-4x+4xy-2y+y^2+6y+9-8\)
\(=\left(2x+y-1\right)^2+\left(y+3\right)^2-8\ge-8\)
=>\(A\ge-4\)
dấu = xảy ra <=> y=-3 và x=2
^_^
\(A=2x^2+2xy+y^2-2x+2y+1\)
\(=\left(x^2+2xy+y^2\right)+2x-4x+2y+1+x^2\)
\(=\left(x+y\right)^2+\left(2x+2y\right)+\left(x^2-4x+1\right)\)
\(=\left(x+y\right)^2+2\left(x+y\right)+\left(x^2-4x+4\right)-4+1\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1+\left(x-2\right)^2-4\)
\(=\left(x+y+1\right)^2+\left(x-2\right)^2-4\ge-4\)
Min A=-4 khi \(\hept{\begin{cases}x+y+1=0\\x-2+0\end{cases}\Rightarrow\hept{\begin{cases}y=-3\\x=2\end{cases}}}\)
cảm ơn các bạn nhiều nhiều nha :))............................
