$A={}^{}-2xy+6{}^{}-12x+2y+45$

$=\left({}^{}-2xy+{}^{}-12x+12y+36\right)+\left(5{}^{}-10y+5\right)+4$

$=\left[{}^{}-12\left(x+y\right)+{}^{}\right]+5\left({}^{}-2y+1\right)+4$

$={}^{}+5{}^{}+4$

Ta có: ${}^{}\ge 0\forall x,y$

$5{}^{}\ge 0\forall y$

$\Rightarrow {}^{}+5{}^{}+4\ge 4\forall x,y$

Dấu "=" xảy ra $\iff x=7,y=1$

Vậy AMIN=4⇔x=7,y=1

đề bài này đúng ko bạn : x² -2xy + 6y²-12x+2y+45

ko đúng bn ơi 

A = x2 - 2xy +6y2 - 12x + 2y +45 

A = (x - y - 6)² - 6y² - 2y - 45 - (y² - 12y - 36)

A = (x - y -6)² + 5(y-1)² +4 \(\ge\)4

Amin = 4 khi y = 1; x = 7

#chanh

a: Ta có: \(B=x^2-4x+6\)

\(=x^2-4x+4+2\)

\(=\left(x-2\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=2

x^2 - 2xy + 6y^2 - 12x + 2y +45 
= x^2 - 2x(y+6) + (y+6)^2 - (y+6)^2 + 6y^2 +2y + 45 
= (x - y - 6)^2 - y^2 - 12y - 36 + 6y^2 + 2y + 45 
= (x - y - 6)^2 + 5y^2 - 10y + 9 
= (x - y - 6)^2 + 5.(y^2 - 2y +1) + 4 
= (x - y - 6)^2 + 5.(y-1)^2 + 4 
=>> MIN = 4 khi (x;y) = {(7;1)}

\(A=x^2-2xy+6y^2-12x+2y+45\)

\(=x^2+y^2+36-2xy-12x+12y+5y^2-10y+5+4\)

\(=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

GTNN A = 4 Khi: \(\hept{\begin{cases}y-1=0\\x-y-6=0\end{cases}\Rightarrow\hept{\begin{cases}y=1\\x=7\end{cases}}}\)

\(A=x^2-2xy+6y^2-12x+2y\)\(+45\)

 \(=x^2+y^2+36-2xy-12x\)\(+12y+5y^2-10y+5+4\)

 \(=\left(x-y-6\right)^2+5\left(y-1\right)^2\)\(+4\ge4\)

GTNN của A là 4 khi \(\hept{\begin{cases}y-1=0\\x-y-6=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}y=1\\x-y=6\end{cases}\Rightarrow\hept{\begin{cases}y=1\\x=7\end{cases}}}\)

Vậy BT A đạt giá trị nhỏ nhất là 4 tại x = 7 và y = 1

4 Ok

\(A=x^2-2xy+6y^2-12x+2y+45\)

\(A=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5+4\)

\(A=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y^2-2y+1\right)+4\)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)

Do : \(\left(x-y-6\right)^2\text{≥}0\) ∀\(xy\) ; \(5\left(y-1\right)^2\text{≥}0\text{∀}y\)

⇒ \(\left(x-y-6\right)^2+5\left(y-1\right)^2\text{ ≥}0\)

⇔ \(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\text{≥}4\)

⇒ \(A_{Min}=4."="\text{⇔}x=7;y=1\)

\(A=x^2-2xy+6y^2-12x+2y+54\)

\(A=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5+4\)

\(A=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y^2-2y+1\right)+4\)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)

Do: \(\left(x-y-6\right)^2\ge0\forall xy\); \(5\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-y-6\right)^2+5\left(y-1\right)^2\ge0\)

\(\Leftrightarrow A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

\(\Rightarrow A_{Min}=4\)

Dấu "=" xảy ra khi \(x=7;y=1\)

\(A=x^2-2xy+6y^2-12x+2y+45\)

\(=x^2+y^2+36-2xy-12x+12y+5y^2-10y+5+4\)

\(=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

Gía trị nhỏ nhất : \(A=4\)Khi \(\hept{\begin{cases}y-1=0\\x-y-6=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\x=7\end{cases}}\)