mày phải k bố ko anh gọi cave đến chịch chết mày

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

Ta có \(A=-x^2+2xy-4y^2+2x+10y-3\) 

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

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

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

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

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=y+1\\y-2=0\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(3,2\right)\)

Vậy \(max_A=10\)

?

\(A=-x^2+2xy-4y^2+2x+10y-8\)

\(=-\left(x^2-2xy+4y^2-2x-10y+8\right)\)

\(=-\left[\left(x-y-1\right)^2+3\left(y-2\right)^2-5\right]\)

\(=5-\left(x-y-1\right)^2-3\left(y-2\right)^2\le5\)

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

Vậy MAX  \(A=5\)khi  \(x=3;\)\(y=2\)

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

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

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

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

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

Vậy \(C_{max}=10\) tại x = 3; y = 2