Lời giải:
$A=x^2+2x+2xy+2y^2+4y+2021$

$=(x^2+2xy+y^2)+2x+y^2+4y+2021$

$=(x+y)^2+2(x+y)+1+(y^2+2y+1)+2019$

$=(x+y+1)^2+(y+1)^2+2019\geq 2019$

Vậy $A_{\min}=2019$ khi $x+y+1=y+1=0$

$\Leftrightarrow (x,y)=(0,-1)$

A=x²+2y²+2xy+2x-4y+2013

=x²+y²+1+2xy+2x+2y+y²-6y+9+2003

=(x+y+1)²+(y-3)²+2003

Min A=2003 tại x=-4;y=3

A= (X²+2XY+Y²) + 2(X+Y)+1+Y²-6Y+9+2003

A=(X+Y)²+ 2(X+Y)+1+(Y-3)²+2003

A=(X+Y+1)²+(Y-3)²+2003

=> A>=2003

(DẤU "=" XẢY RA KHI X=-4;Y=3)

2) ĐKXĐ:  \(1\le x\le5\)

\(B^2=\left(\sqrt{x-1}+\sqrt{5-x}\right)^2\le\left(1^2+1^2\right)\left(x-1+5-x\right)=8\Rightarrow B\le2\sqrt{2}\)

Xảy ra đẳng thức khi và chỉ khi x = 3

\(A=\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(y^2-6y+9\right)+2006\)\(=\left(x+y\right)^2+2\left(x+y\right)+1+\left(y-3\right)^2+2006\)

\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2006\)

Ta có: \(\left(x+y+1\right)^2+\left(y-3\right)^2\ge0\left(\forall x;y\right)\)

\(\Rightarrow A\ge2006\).

Vậy MIN A = 2006 \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)

Bài 2: 

a: \(=-\left(x^2+2x-100\right)\)

\(=-\left(x^2+2x+1-101\right)\)

\(=-\left(x+1\right)^2+101< =101\)

Dấu = xảy ra khi x=-1

b: \(=-3\left(x^2-\dfrac{1}{3}x\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{1}{36}\right)\)

\(=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{1}{12}< =\dfrac{1}{12}\)

Dấu = xảy ra khi x=1/6

c: \(=-\left(3x^2+4y^2-18x+8y-12\right)\)

\(=-\left(3x^2-18x+27+4y^2+8y+4-43\right)\)

\(=-3\left(x-3\right)^2-4\left(y+1\right)^2+43< =43\)

Dấu = xảy ra khi x=3 và y=-1

2P = \(2x^2+4xy+4y^2-12x-8y+50\)

      = \(\left(x+2y\right)^2-2\left(x+2y\right)\cdot2+4+x^2-8x+16+30\)

      = \(\left(x+2y-2\right)^2+\left(x-4\right)^2+30\ge30\)

=> P \(\ge15\)

Dấu '' = '' xảy ra khi x = 4 ; y = -1

P = x² + 2y² + 2xy - 6x - 4y + 25 đạt GTNN khi x² + 2y² + 2xy - 6x - 4y = -25 và P = 0

Lập luận đỉnh cao!! ^~^

19

(ko bít có phải ko)

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

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

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

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

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