a)\(A=5-8x-2x^2\)

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

\(=-2\left(x^2+4x+4-\frac{13}{2}\right)\)

\(=-2\left[\left(x+2\right)^2-\frac{13}{2}\right]\)

\(=-2\left[\left(x+2\right)^2\right]+13\le13\)

Vậy \(A_{max}=13\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

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 \(max_A=10\)

?

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

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

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

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

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

\(B=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-4\left(x^2-2xy+y^2\right)-\left(y^2-10y+25\right)+37\)

\(=-4\left(x-y\right)^2-\left(y-5\right)^2+37< =37\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y=0\\y-5=0\end{matrix}\right.\)

=>x=y=5

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

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

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

Vậy \(MaxA=10\), đạt được khi và chỉ khi \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Lời giải:
$-A=x^2-2xy+4y^2-2x-10y+3$

$=(x^2-2xy+y^2)+3y^2-2x-10y+3$

$=(x-y)^2-2(x-y)+3y^2-12y+3$

$=(x-y)^2-2(x-y)+1+3(y^2-4y+4)-10$

$=(x-y+1)^2+3(y-2)^2-10\geq 0+0-10=-10$

$\Rightarrow A\leq 10$

Vậy $A_{\max}=10$. Giá trị này đạt tại $x-y+1=y-2=0$

$\Leftrightarrow y=2; x=1$

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

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

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

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

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  \(x=3;\)\(y=2\)

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

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

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

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

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

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