\(M=2x^2+5y^2-2xy+2y+2x\)

\(2M=4x^2+10y^2-4xy+4y+4x\)

\(2M=\left(4x^2-4xy+y^2\right)+9y^2+4x+4y\)

\(2M=\left[\left(2x-y\right)^2+2\left(2x-y\right)+1\right]+\left(9y^2+6y+1\right)-2\)

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

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

         \(\left(3y+1\right)^2\ge0\forall y\)

\(\Rightarrow2M\ge-2\)

\(\Leftrightarrow M\ge-1\)

Dấu "=" xảy ra khi :

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

Vậy ....

Mình chưa hiểu ở dòng thứ 3

dòng 3 là hằng đẳng thức bạn ơi

E = 2x^2 - 5x -2 = 2( x^2 -5/2x -1) = 2(x^2 - 2.x.5/4 +25/16 - 41/16) = 2(x - 5/4 )^2 + 41/8

Vậy GTNN của biểu thức là 41/8 tại x = 5/4

F = x^2 + 5y^2 + 2xy -y +3 = (x^2 + 2xy +y^2) + (4y^2 - 2.2y.1/4 + 1/16) +47/16

(x + y)^2  + (2y - 1/4)^2 + 47/16 

Vậy GTNN của BT là 47/16 tại x = y = 1/8

C= \(\frac{1}{2}\)

a) -( x-y)² - (x-1)² -2 

GTLN = -2

=x²-2xy+y²+4y²+4y+1+2

=(x-y)²+(2y+1)²+2\(\ge2\)

dấu bằng xảy ra khi x=y=-1/2

1. \(A=2x^2-6x-2xy+y^2+10\)

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

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

Vì \(\left(x-y\right)^2\ge0\) ; \(\left(x-3\right)^2\ge0\)\(\forall x;y\)

\(\Rightarrow A=\left(x-y\right)^2+\left(x-3\right)^2+1\ge1\)

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

Vậy minA = 1 \(\Leftrightarrow x=y=3\)

2. \(A=5+2xy+14y-x^2-5y^2-2x\)

\(\Leftrightarrow A=-\left(x^2-2xy+y^2+2x-2y+1\right)-\left(4y^2-12y+9\right)+15\)

\(\Leftrightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15\)

Vì \(\left\{{}\begin{matrix}\left(x-y+1\right)^2\ge0\\\left(2y-3\right)^2\ge0\end{matrix}\right.\)\(\forall x;y\)

\(\Rightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15\le15\)

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

Vậy maxA = 15 \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{3}{2}\end{matrix}\right.\)

1.

Vì ;

Dấu "=" xảy ra

Vậy minA = 1

2.

Vì

Dấu "=" xảy ra

Vậy maxA = 15

có làm mới có ăn nha em

\(M=5x^2+y^2-2x+2y+2xy+2004\)

\(=\left(x^2+2x+1\right)+2y\left(x+1\right)+y^2+4x^2-4x+1+2002\)

\(=\left(x+1\right)^2+2y\left(x+1\right)+y^2+\left(2x-1\right)^2+2002\)

\(=\left(x+1+y\right)^2+\left(2x-1\right)^2+2003\ge2002\) với mọi x,y

=> \(M_{min}=2002\Leftrightarrow\left\{{}\begin{matrix}x+y+1=0\\2x-1=0\end{matrix}\right.\)

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

Vậy \(M_{min}=2002\)