\(A=x^2+2x\left(y+1\right)+\left(y+1\right)^2-\left(y+1\right)^2+2y^2-4y+2028\)

\(=\left(x+y+1\right)^2-y^2-2x-1+2y^2-4y+2028\)

\(=\left(x+y+1\right)^2-6x+y^2+2027\)

\(=\left(x+y+1\right)+\left(y-3\right)^2+2018\ge2018\forall x;y\) (do...)

=> MinA = 2018 \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)

Ta có : \(x^2+y^2-2x+4y+1\)

\(=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)-4\)

\(A=\left(x-1\right)^2+\left(y+2\right)^2-4\)

Vì \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\in R\)

Nên : \(A=\left(x-1\right)^2+\left(y+2\right)^2-4\ge-4\forall x,y\in R\)

Vậy \(A_{min}=-4\) khi x = 1 và y = -2

\(A=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(y^2-6y+9\right)+2018\)

\(A=\left(x+y+1\right)^2+\left(y-3\right)^2+2018\ge2018\)

\(A_{min}=2018\) khi \(\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)

Giúp mk bài hình mk mới đăng với Nguyễn Việt Lâm Quản lý, ý b,c, d thôi

biet tong cua so thu nhat va so thu hai bang 5,8.Tong cua so thu hai va so thu ba bang 6,7.Tong so thu nhat va so thu ba bang 7,5.Tim moi so do?

\(A=x^2+2xy+2y^2+2x-4y+2013\)

\(=\left(x^2+y^2+1+2x+2y+2xy\right)-1-2y+y^2-4y+2013\)\(=\left(x+y+1\right)^2+\left(y^2-2.y.3+9\right)-9+2012\)

\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2003\)

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

\(\Rightarrow A=x^2+2xy+2y^2+2x-4y+2013=\left(x+y+1\right)^2+\left(y-3\right)^2+2003\ge2003\)

\(\Rightarrow Min\left(A\right)=2003\)

còn thiếu: khi y=3 và x= -y-1

\(=\left(x^2+4x+4\right)+\left(y^2+4y+4\right)+\left(x^2-2xy+y^2\right)+2=\left(x+2\right)^2+\left(y+2\right)^2+\left(x-y\right)^2+2\ge2\)

=> Min =2 <=> x=y=-2