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 \(max_A=10\)

?

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

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

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

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

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  \(x=3;\)\(y=2\)

mày phải k bố ko anh gọi cave đến chịch chết mày

\(C=-3x\left(3+x\right)-7=-9x-3x^2-7=-\left(3x^2+9x+7\right)=-3\left(x^2+3x+\frac{7}{3}\right)\)

=\(-3\left(x^2+2.\frac{3}{2}.x+\frac{9}{4}+\frac{1}{12}\right)=-3\left[\left(x+\frac{3}{2}\right)^2+\frac{1}{12}\right]=-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\le-\frac{1}{4}\)

Dấu "=" xảy ra khi x=-3/2

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

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

\(=-\left[x^2+2x\left(y-1\right)+4y^2-10y+8\right]\)

\(=-\left[x^2+2x\left(y-1\right)+\left(y^2-2y+1\right)+3y^2-8y+7\right]\)

\(=-\left[x^2+2x\left(y-1\right)+\left(y-1\right)^2+3\left(y^2-2.\frac{4}{3}.y+\frac{16}{9}\right)+\frac{5}{3}\right]\)

\(=-\left[\left(x+y-1\right)^2+3\left(y-\frac{4}{3}\right)^2+\frac{5}{3}\right]\)

\(=-\left(x+y-1\right)^2-3\left(y-\frac{4}{3}\right)^2-\frac{5}{3}\le-\frac{5}{3}\)

Dấu "=" xảy ra khi x=-1/3 và y=4/3

-A= x^2-2xy+4y^2-2x-10y+8 
-A= ( x^2+y^2+1-2xy-2x+2y) +(3y^2-12y+7) 
-A=(x-y-1)^2+ 3(y^2-4y+7/4)=(x-y-1)^2+3(y-2)^2-27/4>=-... nen A<= 27/4 

(ko biết có đúng hay ko)

-A= x^2-2xy+4y^2-2x-10y+8 
-A= ( x^2+y^2+1-2xy-2x+2y) +(3y^2-12y+7) 
-A=(x-y-1)^2+ 3(y^2-4y+7/4)=(x-y-1)^2+3(y-2)^2-27/4>=-... nen A<= 27/4 

tích đúng mình giải cho