E = x^2 + y^2 + 2xy + x^2 - 8x + 16 + 2012

=> E = (x + y)^2 + (x - 4)^2 + 2012

=> E nhỏ nhất bằng 2012 <=> x = 4 ; y = -4 

B=(x²+2,x,y+y²)+(x²-2.x.4+4²)+2012

B=(x+y)²+(x-4)²+2012

   (x+y)² lớn hoăc bằng 0 (mình ko ghi dc ki hiệu)

  (x-4)² lớn hoăc bằng 0 (mình ko ghi dc ki hiệu)

=>(x+y)²+(x-4)²+2012 lớn hoăc bằng 2012

Dấu = xảy ra khi x+y=0 => x=-4

                          x-4=0 => x=4

a, Tìm GTNN

\(A=2x^2+y^2+2xy-8x+2028\)

\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+16\right)+2012\)

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

Ta có :

\(\left(x+y\right)^2\ge0\) với mọi x

\(\left(x-4\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x+y\right)^2+\left(x-4\right)^2+2012\ge2012\)

Dấu = xảy ra

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)^2=0\\\left(x+y\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\x+y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-4\end{matrix}\right.\)

Vậy \(Min_A=2012\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-4\end{matrix}\right.\)

A=2x²+y²+2xy-8x+2028=(x²+2xy+y²)+(x²-8x+16)+2012=(x+y)²+(x-4)²+2012

Vì (x+y)^{2\(\ge\)0\(\forall\)x,y}

(x-4)^{2\(\ge0\forall x\)}

^=>(x+y)²+(x-4)²\(\ge0\)

=>(x+y)²+(x-4)²+2012\(\ge2012\forall x,y\)

Đạt được khi và chỉ khi:

\(\left\{{}\begin{matrix}x-4=0\rightarrow x=4\\x+y=0\rightarrow y=-4\end{matrix}\right.\)

Vậy Amin=2012<=>x=4,y=-4

a) A=2x²+y²+2xy-8x+2028

=(x²+2xy+y²)+(x²-8x+16)+2012

=(x+y)²+(x-4)²+2012

do (x+y) ²≥ 0 ∀x;y

(x-4)²≥ 0 ∀x

=> (x+y)²+(x-4)² ≥ 0

=> (x+y)²+(x-4)²+2012 ≥ 2012

=> A≥2012

vậy GTNN A=2012 khi \(\left[{}\begin{matrix}x+y=0\\x-4=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}y=-4\\x=4\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

\(B=2x^2+y^2-8x+2xy-4y+2025\)

\(=x^2+2xy+y^2-4x-4y+x^2-4x+2025\)

\(=\left(x+y\right)^2-4\left(x+y\right)+4+x^2-4x+4+2017\)

\(=\left(x+y-2\right)^2+\left(x-2\right)^2+2017\ge2017\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-2=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=-x+2=-2+2=0\end{cases}\)

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

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

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