\(A=-2x^2-10y+4xy+4x+4y+2013\)

\(A=-\left(2x^2+10y^2-4xy-4x-4y-2013\right)\)

\(A=-\left(x^2+x^2+y^2+9y^2+2xy-6xy-4x-4y-2013\right)\)

\(A=-\left[\left(x^2+2xy+y^2\right)-4\left(x+y\right)+4+\left(3y\right)^2-2\cdot3y\cdot x+x^2-2017\right]\)

\(A=-\left[\left(x+y\right)^2-2\cdot\left(x+y\right)\cdot2+2^2+\left(3y-x\right)^2-2017\right]\)

\(A=-\left[\left(x+y\right)^2+\left(3y-x\right)^2-2017\right]\)

\(A=2017-\left(x+y\right)^2-\left(3y-x\right)^2\)

\(A=2017-\left[\left(x+y\right)^2-\left(3y-x\right)^2\right]\le2017\)

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

\(A=-2x^2-10y^2+4xy+4x+4y+2013\)

   \(=-2\left(x-y\right)^2+4\left(x-y\right)-2-8y^2+8y-2+2017\)

   \(=-2\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]-8\left(y^2-y+\frac{1}{4}\right)+2017\)        

   \(=-2\left(x-y-1\right)^2-8\left(y-\frac{1}{2}\right)^2+2017\le2017\forall x;y\)

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

Vậy GTLN của A là 2017 khi \(x=\frac{3}{2}\)và \(y=\frac{1}{2}\)

Chúc bạn học tốt.

Sử dụng các hằng đẳng thức: (a-b-c)²=a^2+b^2+c^2-2ab-2ac+2bc

A= -2(x²+y²-2xy-2x+2y+1)-8y²+8y+2+2013=-2(x-y-1)²-8(y²-2.y.1/2+1/4)+2+2+2013=-(x-y-1)²-(y-1/2)²+2017\(\le2017\)

'=' xảy ra khi và chỉ khi \(\hept{\begin{cases}x-y-1=0\\y-\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}}\)

Vậy gtln của A=2017 khi x=3/2 và y=1/2

\(A=-2x^2+4xy-2y^2+4\left(x-y\right)-2-8y^2+8y+2019\\ A=\left[-2\left(x-y\right)^2+4\left(x-y\right)-2\right]-8\left(y^2-y+\dfrac{1}{4}\right)+2020\\ A=-2\left(x-y-1\right)^2-8\left(y-\dfrac{1}{2}\right)^2+2020\le2020\\ A_{max}=2020\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1+\dfrac{1}{2}=\dfrac{3}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Cịu truyền nghề cho cháu với ạ!

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

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

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

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

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

$\Leftrightarrow y=2; x=1$