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

\(A=-2\left(x^2+y^2+1-2xy-2x+2y\right)-2\left(4y^2-4y+1\right)+2017\)

\(A=-2\left(x-y-1\right)^2-2\left(2y-1\right)^2+2017\le2017\)

\(A_{max}=2017\) khi \(\left\{{}\begin{matrix}x=\frac{3}{2}\\y=\frac{1}{2}\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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.

Sửa dòng 6 :

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

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y-2=0\\3y-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=2\\3y=x\end{cases}\Leftrightarrow}\hept{\begin{cases}4y=2\\x+y=2\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{1}{2}\\x=\frac{3}{2}\end{cases}}}\)

P.s : bài bạn Pham Van Hung đúng rồi, đây là một cách khác nhé

\(A=-2x^2-10y^2+4xy+4x+4y+2016\)

\(=-2.\left(x^2+5y^2-4xy-4x-4y\right)+2016\)

\(=-2.\left(x^2+4y^2+4-4xy-4x+8y+y^2-12y+36\right)+2.36+2016\)

\(=-2.[\left(x-2y-2\right)^2+\left(y-6\right)^2]+2088\)

Ta có: \(\left(x-2y-2\right)^2+\left(y-6\right)^2\ge0\)

\(\Rightarrow-2.[\left(x-2y-2\right)^2+\left(y-6\right)^2]\le0\)

\(\Rightarrow-2.[\left(x-2y-2\right)^2+\left(y-6\right)^2]+2088\le2088\)

\(\Rightarrow A\le2088\)

Vậy giá trị lớn nhất của \(A=2088\) khi: \(\hept{\begin{cases}x-2y-2=0\\y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=2y+2\\y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=14\\y=6\end{cases}}\)

\(A=-2x^2-10y^2+4xy+4x+4y+2016\\ A=-2x^2+4xy-4y^2+4\left(x-y\right)-2-6y^2+8y+2018\\ A=-2\left(x-y\right)^2+4\left(x-y\right)-2-6\left(y^2-\dfrac{4}{3}y\right)+2018\\ A=-2\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]-6\left(y^2-2\cdot\dfrac{2}{3}y+\dfrac{9}{4}\right)+\dfrac{27}{2}+2018\\ A=-2\left(x-y-1\right)^2-6\left(y-\dfrac{3}{2}\right)^2+\dfrac{4063}{2}\le\dfrac{4063}{3}\\ A_{max}=\dfrac{4063}{2}\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\y=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\)