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

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

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

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

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  \(x=3;\)\(y=2\)

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.

ta có:

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

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

=(x-y)^2-2(x-y)+1+3(y^2-4y+4)-(13-8)

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

vì (x-y-1)^2\(\ge0\)với mọi x,y

3(y-2)^2\(\ge0\)với mọi y

suy ra (x-y-1)^2+3(y-2)^2-5\(\ge-5\)với mọi x,y

dấu "=" xảy ra\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\y=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\)

Vậy GTNN của M là -5 khi \(\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\)

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

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

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

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

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

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

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 : \(5\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

\(-x^2+2xy-4y^2+2x+10y-8=-x^2+2x\left(y+1\right)-\left(y+1\right)^2-3y^2+12y-7\)

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

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

Vậy B đạt giá trị lớn nhất bằng 5 tại (x;y) = (3;2)