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

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

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

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

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 = x² -2xy + 2y²+ 2x - 10y -5

= x² - 2xy + y² + y² + 2x - 2y - 8y -5

= [(x² - 2xy + y²) + 2 ( x - y) + 1]² + (y² - 8y + 16) - 22     

= [ (x - y)² + 2(x - y) + 1]² + (y - 4)²  - 22

= (x - y + 1)² + ( y - 4)² - 22 ≥ -22

=> Min của A = -22 khi {y−4=0x−y+1=0{y−4=0x−y+1=0 => {y=4x−3=0{y=4x−3=0 => {y=4x=3{y=4x=3

Vậy Min của A = 2016 khi x = 3 và y = 4.

MinA=-22 khi \(\hept{\begin{cases}\left(y-4\right)^2=0\\\left(x-y+1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=4\\x=3\end{cases}}}\)

\(A=x^2-2xy+4y^2-2x-10y-5\)

\(A=x^2-2x\left(y+1\right)+4y^2-10y-5\)

\(A=x^2-2x\left(y+1\right)+4\left(y^2+2y+1\right)-18y-9\)

\(A=x^2-2x\left(y+1\right)+\left(y+1\right)^2+3\left(y^2+2y+1\right)-18y-9\)

\(A=\left(x-y-1\right)^2+3y^2-12y-6\)

\(A=\left(x-y-1\right)^2+3\left(y^2-4y+4\right)-18\)

\(A=\left(x-y-1\right)^2+3\left(y-2\right)^2-18\)

Thấy: \(\left(x-y-1\right)^2\ge0;3\left(y-2\right)\ge0\forall x;y\)

\(\Rightarrow A\ge-18\). Vậy \(Min_A=-18\)

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

Giups mik vs

Do A nhỏ nhất 

Suy ra : x^2 = 0, 2y^2 = 0 , 4y = 0 .......( tất cả số hạng bằng 0) 

Suy ra A= 2019

\(A=x^2+2y^2+4y+2xy-4x+2019\)

\(A=\left(x^2+y^2-2^2+2xy-4y-4x\right)+\left(y^2+8y+4^2\right)+2007\)

\(A=\left(x+y-2\right)^2+\left(y+4\right)^2+2007\ge2007\)

Vậy \(Min_A=2007\) khi \(\hept{\begin{cases}x+y-2=0\\y+4=0\end{cases}}\hept{\begin{cases}x+y=2\\y=-4\end{cases}}\hept{\begin{cases}x=6\\y=4\end{cases}}\)