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

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

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

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

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

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

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

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

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

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?

mik o biet nhung k nha minh nhanh nhat

a,   B=x²+4xy+y²+x²-8x+16+2012

       B=(x+y) ²+(x-4)²+2012

 Vậy B >=2012 ( Dấu "=" xảy ra khi x=4,y=-4)

b làm tương tự 

c,  9x²+6x+1+y²-4y+4+x²-4xz+4z²=0

     (3x+1)²+(y-4)²+(x-2z)²=0

    Vậy 3x+1=0 => x = -1/3

           y-4=0 => y=4

             x-2z=0  thế x=-1/3 ta được.      -1/3-2z=0 => z = -1/6

Bạn nhớ ghi lại đề minh không ghi đề 

a) \(B=2x^2+y^2+2xy-8x+2028\)

\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+4^2\right)+2012=\left(x+y\right)^2+\left(x-4\right)^2+2012\ge2012\)

\(MinB=2012\Leftrightarrow\hept{\begin{cases}x=4\\y=-4\end{cases}}\)

b)\(C=x^2+5y^2+4xy+2x+2y-7\)

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

\(=\left(\left(x+2y\right)^2+2\left(x+2y\right)+1\right)+\left(y-1\right)^2-9=\left(x+2y+1\right)^2+\left(y-1\right)^2-9\ge9\)

\(MinC=-9\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

c)\(10x^2+y^2+4z^2+6x-4y-4xz+5=0\)

\(\Leftrightarrow\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)^2+\left(y-2\right)^2+\left(x-2z\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}\)