Ta có: 

Vậy giá trị lớn nhất của  khi: \hept{x−2y−2=0y=6⇒\hept{x=2y+2y=6⇒\hept{x=14y=6\hept{x−2y−2=0y=6⇒\hept{x=2y+2y=6⇒\hept{x=14y=6

Thu gọn

\(A=-2\left(x^2+2xy+y^2\right)+4\left(x+y\right)-2-8y^2+2018\\ A=-2\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]-8y^2+2018\\ A=-2\left(x+y-1\right)^2-8y^2+2018\le2018\\ A_{max}=2018\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

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

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

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

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

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Biểu thức gì vậy bạn