\(A=x^2+2y^2+2xy+2x-4y+2016\)

\(=x^2+y^2+y^2+2xy+2x+2y-6y+2016\)

\(=\left(x^2+2xy+y^2\right)+\left(y^2-6y+9\right)+\left(2x+2y\right)+2007\)

\(=\left(x+y\right)^2+\left(y-3\right)^2+2\left(x+y\right)+2007\)

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

Vì \(\hept{\begin{cases}\left(x+y+1\right)^2\ge0;\forall x,y\\\left(y-3\right)^2\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(x+y+1\right)^2+\left(y-3\right)^2\ge0;\forall x,y\)

\(\Rightarrow\left(x+y+1\right)^2+\left(y-3\right)^2+2006\ge0+2006;\forall x,y\)

Hay \(A\ge2006;\forall x,y\)

Dấu"=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Vậy \(A_{min}=2006\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Mình làm có gì sai hả @@ 

do em điểm cao qua mà

tích cho a đi

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)

Đặt `A=2x^2+2y^2+2xy-4x+4y+2021`

`<=>2A=4x^2+4y^2+4xy-8x+8y+4042`

`<=>2A=4x^2+4xy+y^2-8x-4y+3y^2+12y+4042`

`<=>2A=(2x+y)^2-4(2x+y)+4+3y^2+12y+12+4026`

`<=>2A=(2x+y-2)^2+3(y+2)^2+4026>=4026`

`=>A>=2013`

Dấu "=" xảy ra khi `y=-2,x=(2-y)/2=2`

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

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

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

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

D ez nhất :v

\(D=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+5\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+5\ge5\)

Đẳng thức xảy ra khi x = 1 và y = -2

\(A=\left[\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4\right]+\left(y^2-2y+1\right)+2020\)

\(=\left[\left(x-y\right)^2+2\left(x-y\right).2+2^2\right]+\left(y-1\right)^2+2020\)

\(=\left(x-y+2\right)^2+\left(y-1\right)^2+2020\ge2020\)

Dấu "=" xảy ra khi y = 1 và x - y + 2 = 0 tức là x = y - 2 = -1

\(B=\left(x^2-2xy+y^2\right)-2\left(x-y\right)+1+x^2-2x+1+2019\)

\(=\left(x-y\right)^2-2\left(x-y\right).1+1+\left(x-1\right)^2+2019\)

\(=\left(x-y-1\right)^2+\left(x-1\right)^2+2019\ge2019\)

Dấu "=" xảy ra khi x = 1 và x - y - 1 = 0 hay y = 0

M = 5 - x² + 2x - 4y² - 4y

= (- x² + 2x - 1) + (- 4y² - 4y - 1) + 7

= 7 - (x - 1)² - (2y + 1)²\(\le7\)

Dấu "=" xảy ra khi x = 1 và y = - 0,5

(^~^)

M = - x² + 2xy - 4y² + 2x + 10y - 8

- M = x² - 2xy + 4y² - 2x - 10y + 8

= (y² + 1 + x² + 2y - 2xy - 2x) + (3y^2 - 12y + 12) - 5

\(=\left(y+1-x\right)^2+3\left(y-2\right)^2-5\ge-5\)

\(\Rightarrow M\le5\)

Dấu "=" xảy ra khi y = 2 và x = 3.

\(C=2\left(x-\frac{5}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\Rightarrow C_{min}=\frac{7}{8}\)

\(D=\left(x^2+4xy+4y^2\right)+\left(y^2+y+\frac{1}{4}\right)+\frac{8083}{4}\)

\(D=\left(x+2y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{8083}{4}\ge\frac{8083}{4}\)

\(E=\frac{1}{2}\left(4x^2+y^2+\frac{9}{4}-4xy-6x+3y\right)+\frac{1}{2}\left(y^2+y+\frac{1}{4}\right)+\frac{15}{4}\)

\(E=\frac{1}{2}\left(2x-y-\frac{3}{2}\right)^2+\frac{1}{2}\left(y+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\)

\(A=-\left(x-2\right)^2+11\le11\)

\(B=-\left(x+\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)

\(C=-\left(x-3y\right)^2-\left(y-2\right)^2+11\le11\)