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

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

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

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

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

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

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

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

P = x² + 2y² + 2xy – 6x – 8y + 2028

P = (x² + y² + 2xy) – 6(x + y) + 9 + y² – 2y + 1 + 2018

P = (x + y – 3)² + (y – 1)² + 2018 \(\ge\) 2018

=> Giá trị nhỏ nhất của P = 2018 khi x = 2; y = 1

P=x²+2y²+2xy-6x-8y+2028

=x²+2xy+y²+y²-8y+x²-6x-x²+2028

=(x²+2xy+y²)+(y²-8y+16)+(x²-6x+9)-x²+2028-16-9

=(x-y)²+(y-4)²+(x-3)²-x²+2003\(\ge2003\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-4\right)^2\ge0\\\left(x-3\right)^2\ge0\\x^2\ge0\end{matrix}\right.\) nên:

Để P=2003 thì :

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-3\right)^2=0\\\left(y-4\right)^2=0\\x^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-3=0\\y-4=0\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=4\\x=0\end{matrix}\right.\)

Vậy min P=2003\(\Leftrightarrow\left(x=y\right)\in\left\{0;4;3\right\}\)

P = x² + 2y² + 2xy – 6x – 8y + 2028

P = (x² + y² + 2xy) – 6(x + y) + 9 + y² – 2y + 1 + 2018

P = (x + y – 3)² + (y – 1)² + 2018  2018

=> Giá trị nhỏ nhất của P = 2018 khi x = 2; y = 1

Lời giải:

$P=(x^2+y^2+2xy)+y^2-6x-8y+2028$

$=(x+y)^2-6(x+y)+(y^2-2y)+2028$
$=(x+y)^2-6(x+y)+9+(y^2-2y+1)+2018$

$=(x+y-3)^2+(y-1)^2+2018\geq 0+0+2018=2018$

Vậy $P_{\min}=2018$

Giá trị này đạt tại $x+y-3=y-1=0$

$\Leftrightarrow y=1; x=2$

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

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

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

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

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 

Giải :(x²+2xy+y²)+y²-6x-8y+2024=(x+y)²-2(x+y)3+y²-2y+2024

=(x+y-3)²+(y²-2y+1)+2014=(x+y-3)²+(y-1)²+2014 >=2014

vì (x+y-3)²;(y-1)²>=0 với mọi x;y

nên Pmin=2014khi y=1;x=2

MinP=2024 nha!

2024 đó !đúng 100% luôn !