\(F=\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+2021\\ F=\left(x-y\right)^2+\left(y-1\right)^2+2021\ge2021\)

Dấu \("="\Leftrightarrow x=y=1\)

Vậy \(F_{min}=2021\)

\(\Rightarrow F=\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+2021\\ \Rightarrow F=\left(x-y\right)^2+\left(y-1\right)^2+2021\ge2021\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

A= x²+2y²-2xy-2x-2y+1015

A = x² - 2xy - 2x + y² + 2y + 1 + y² - 4y + 4 + 1010 

A = [x² - 2x(y + 1) + (y+1)² ]  + (y-2)² + 1010

A = ( x - y - 1)² + (y-2)² + 1010 \(\ge1010\forall x,y\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\)

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

Vậy MinA = 1010 <=> \(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Bạn nên sửa lại đề là tìm GTNN

\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+y^2+4y+4+15\\ A=\left(x-y+1\right)^2+\left(y+2\right)^2+15\ge15\\ A_{min}=15\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

Vậy GTNN của A là 15

\(Q=x^2+2y^2+2z^2+2xy-2yz-2xz-2y+4z+5=\left[\left(x^2+2xy+y^2\right)-2z\left(x+y\right)+z^2\right]+\left(y^2-2y+1\right)+\left(z^2+4z+4\right)=\left(x+y-z\right)^2+\left(y-1\right)^2+\left(z+2\right)^2\ge0\)

\(minQ=0\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-3\\y=1\\z=-2\end{matrix}\right.\)

`Q=x^2+2y^2+2z^2+2xy-2yz-2xz-2y+4z+5`

`Q=(x^2+y^2-z^2+2xy-2yz-2xz)+(y^2-2y+1)+(z^2+4z+4)`

`Q=(x+y-z)^2+(y-1)^2+(z+2)^2`

Ta thấy :

`(x+y-z)^2>=0`

`(y-1)^2>=0`

`(z+2)^2>=0`

`=>(x+y-z)^2+(y-1)^2+(z+2)^2>=0`

Dấu = xảy ra 

`<=>` $\begin{cases}x+y-z=0\\y-1=0\\z+2=0\end{cases}$

`<=>` $\begin{cases}x=-3\\y=1\\z=-2\end{cases}$

=>x^2-2xy+y^2+y^2+2y+1=0

=>(x-y)^2+(y+1)^2=0

=>x=y=-1

B=-2022-2023=-4045

Lời giải:

a. $x^2+y^2+4y+13-6x$

$=(x^2-6x+9)+(y^2+4y+4)$

$=(x-3)^2+(y+2)^2$

b.

$4x^2-4xy+1+2y^2-2y$

$=(4x^2-4xy+y^2)+(y^2-2y+1)$

$=(2x-y)^2+(y-1)^2$

c.

$x^2-2xy+2y^2+2y+1$

$=(x^2-2xy+y^2)+(y^2+2y+1)$

$=(x-y)^2+(y+1)^2$

a. \(x^2+y^2+4y+12-6x=\left(x^2-6x+9\right)+\left(y^2+4y+4\right)=\left(x-3\right)^2+\left(y+2\right)^2\)b. \(4x^2-4xy+1+2y^2-2y=\left(4x^2-4xy+y^2\right)+\left(y^2-2y+1\right)=\left(2x-y\right)^2+\left(y-1\right)^2\)c. \(x^2-2xy+2y^2+2y+1=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)

a: \(x^2-6x+y^2+4y+13\)

\(=x^2-6x+9+y^2+4y+4\)

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

b: \(4x^2-4xy+1+2y^2-2y\)

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

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

c: \(x^2-2xy+2y^2+2y+1\)

\(=x^2-2xy+y^2+y^2+2y+1\)

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

\(x^2+2y^2+2xy+6x+2y+2027\)

\(=x^2+2x\left(y+3\right)+\left(y+3\right)^2+\left(y^2-4y+4\right)+2014\)

\(=\left(x+y+3\right)^2+\left(y-2\right)^2+2014\)

Ta có: \(\left\{{}\begin{matrix}\left(x+y+3\right)^2\ge0\forall x;y\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.\)\(\Leftrightarrow\)\(\Rightarrow\left(x+y+3\right)^2+\left(y-2\right)^2+2014\ge2014\)\(\forall x;y\)

Dấu " = " xảy ra < = > \(\left\{{}\begin{matrix}\left(x+y+3\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y+3=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-5\end{matrix}\right.\)

x² - 2xy + 2y² - 2x + 6y + 13 = 0 

<=> x² - 2x(y + 1) + 2y² + 6y + 13 = 0 

<=> x² - 2x(y + 1) + (y + 1)² + y² + 4y + 12 = 0 

<=> (x - y - 1)² + (y + 1)² + (y + 2)² + 8 = 0 

Vô lí do VT > 0 vs mọi x; y 

=> Ko tìm đc gtri của N

khongg lam maa ddoi co an thi an cai lon me may