a=(2x+y)^2+(x-1)^2+(y+2)^2+2021-5=2016

Amin=2016

\(A=3x^2+4xy+4y^2-3x-2y+15\)

\(=\left(x^2+4xy+4y^2\right)-\left(x+2y\right)+\frac{1}{4}+2x^2-2x+\frac{59}{4}\)

\(=\left(x+2y-\frac{1}{2}\right)^2+2\left(x-\frac{1}{2}\right)^2+\frac{57}{4}\ge\frac{57}{4}\)

Đẳng thức xảy ra khi x =1/2; y =0

Vậy..

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

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`

\(A=x^2-4xy+2x-4y+3+4y^2\)

\(A=x^2-2.2xy+\left(2y\right)^2+2x-4y+3\)

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

\(A=\left(x-2y-1\right)^2+2\ge2\)

Vậy GTNN của A=2.

đề thật như vậy sao

K=(4x^2+4xy+y^2)+(x^2-2x+1)+(y^2+4y+4)+2016

=(2x+y)^2+(x-1)^2+(y+2)^2+2016 > =2016 với mọi x,y

minK=2016,dấu "=" xảy ra <=> x=1;y=-2

Lát nữa mới sửa đề hả

a, \(P=2x^2+5y^2+4xy+8x-4y+15\)

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

Dấu "="xảy ra khi:\(\hept{\begin{cases}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-4\\y=2\end{cases}}\)

Vậy...

b, \(C=2x^2+4xy+4y^2-3x-1\)

\(=\left(x+2y\right)^2+\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

sau đó giải tương tự câu a nhé

\(A=5x^2+2y^2+4xy-2x+4y+2021\)

\(=4x^2+4xy+y^2+y^2+4y+4+x^2-2x+1+2016\)

\(=\left(2x+y\right)^2+\left(y+2\right)^2+\left(x-1\right)^2+2016\ge2016\)

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

Vậy Min_A=2016 khi \(\left\{\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

thử sức cùng toan10

= (2x +y)² + (x-1)² +(y+2)² +2012 - 1-4

GTNN = 2007