\(A=x-x^2=-\left(x^2-2\times x\times\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)

\(\left(x-\frac{1}{2}\right)^2\ge0\)

\(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\le\frac{1}{4}\)

Vậy Max A = \(\frac{1}{4}\) khi x = \(\frac{1}{2}\)

***

\(B=5-8x-x^2=-\left(x^2+2\times x\times4+4^2-4^2-5\right)=-\left[\left(x+4\right)^2-21\right]\)

\(\left(x+4\right)^2\ge0\)

\(\left(x+4\right)^2-21\ge-21\)

\(-\left[\left(x+4\right)^2-21\right]\le21\)

Vậy Max B = 21 khi x = - 4 

***

\(C=5-x^2+2x-4y^2-4y=-\left(x^2-2\times x\times1+1^2-1^2+\left(2y\right)^2-2\times2y\times1+1^2-1^2-5\right)=-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\)

\(\left(x-1\right)^2\ge0\)

\(\left(2y-1\right)^2\ge0\)

\(\left(x-1\right)^2+\left(2y-1\right)^2-7\ge-7\)

\(-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\le7\)

Vậy Max C = 7 khi x = 1 và y = \(\frac{1}{2}\)

B=-x^2+2xy-4y^2+2x+10y-8
B = (-x^2 - y^2 - 1 + 2xy + 2x - 2y) + (-3y^2 + 12y - 12) + 5
B = -(x^2+y^2+1 - 2xy - 2x + 2y) - 3(y^2 - 4y + 4) + 5
B = - (x - y - 1)^2 - 3(y - 2)^2 +5  5
 Max B = 5 khi x = 3, y = 2

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

nhok lạnh lùng sai oy (3y2-12y+7) =3(y^2-4y+7/3) ri nè

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

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

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

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 $x-y+1=y-2=0$

$\Leftrightarrow y=2; x=1$