B = x²y²+2x²+24xy+16x+191 = [ (xy)^2 + 24xy + 144] + \(\left[\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.4\sqrt{2}+32\right]\)+15

= (xy+12)^2 +(\(\sqrt{2}x\)+\(4\sqrt{2}\))^2 + 15 

( ở đây mik làm tắt) => Min B = 15 khi \(\sqrt{2}x+4\sqrt{2}=0=>x=-4\)và xy+12 = 0 => -4y = -12= > y=3

A= 2x^2+9y^2-6xy-6x-12y+2004

A = (x^2 -6xy +9y^2) + 4(x -3y) + x^2 - 10x + 2004

A = [(x -3y)^2 +4(x -3y) + 4] + (x^2 -10x +25) + 1975

A= (x -3y +2)^2 + (x -5)^2 + 1975

( mik rút mấy cái bước (x-3y+2)^2 = 0, bn làm thì nên thêm vào=> Min A = 1975 vs x= 5 và y = 7/3

D=-x^2+2xy-4y^2+2x+10y-8

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

D = -(x^2+y^2+1 - 2xy - 2x + 2y) - 3(y^2 - 4y + 4) + 5

D= - (x - y - 1)^2 - 3(y - 2)^2 +5 

=> Max D = 5 khi x= 3 và y=2

C = x^2 + y^2 + 9z^2 - 2x + 12y + 6z + 24

C = (x^2-2x+1)+(y^2 + 2y.6 + 36)+ [ (3z)^2 + 2.3z + 1] - 14

C= (x-1)^2 + (y+6)^2 + (3z+1)^2 - 14

....

mà chép lại đề câu B cho mik vs

\(A=-x^2+2xy-4y^2+2x+10y-8\)

\(=-\left(x^2-2xy+4y^2-2x-10y+8\right)\)

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

\(=5-\left(x-y-1\right)^2-3\left(y-2\right)^2\le5\)

Dấu"=" xảy ra  <=>  \(\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}}\) <=>  \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)

Vậy MAX  \(A=5\)khi  \(x=3;\)\(y=2\)

\(A=-x^2+2xy-4y^2+2x+10y-8\)

\(=-x^2+2xy-y^2-3y^2+2x-2y+12y-12+4\)

\(=-\left(x^2-2xy+y^2\right)+\left(2x-2y\right)-1-\left(3y^2-12y+12\right)+5\)

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

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

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

\(A_{max}=5\Leftrightarrow\hept{\begin{cases}\left(x-y-1\right)^2=0\\3\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}}}\)

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

\(KL:A_{max}=5\Leftrightarrow x=3;y=2\)

Ta có \(A=-x^2+2xy-4y^2+2x+10y-3\) 

\(A=-x^2+2\left(y+1\right)x-4y^2+10y-3\)

\(A=-x^2+2\left(y+1\right)x-\left(y+1\right)^2-3y^2+12y-2\)

\(A=-\left[x-\left(y+1\right)\right]^2-3\left(y^2-4y+4\right)+10\)

\(A=-\left(x-\left(y+1\right)\right)^2-3\left(y-2\right)^2+10\) \(\le10\)

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

Vậy \(max_A=10\)

?

\(C=-x^2+2xy-4y^2+2x+10y-3\)

\(=-\left(x^2+2xy-y^2\right)+2x-2y-1-3y^2+12y-12+10\)

\(=-\left(x-y\right)^2+2\left(x-y\right)-1-3\left(y^2-4y+4\right)+10\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+10\le10\forall x;y\)

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

Vậy \(C_{max}=10\) tại x = 3; y = 2

\(-x^2+2xy-4y^2+2x+10y-8\)

\(=-\left(x^2-2xy+y^2\right)+2\left(x-y\right)+12y-8-3y^2\)

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

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

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

\(=-\left[\left(x-y-1\right)^2+3\left(y-2\right)^2\right]+5\le5\forall x;y\)

Dấu " = " xảy ra

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

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

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

Vậy GTLN của biểu thức trên là : \(5\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)