\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right)\\ \)

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

Amin=4 khi y=1; x=7

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

\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right) \)

\(A=\left(x-7-6\right)^2+5\left(y-1^2\right)+4\ge4\)

\(Amin=4\)\(khi\)\(y=1;x=7\)

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

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

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

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

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 \(\hept{\begin{cases}y-1=0\\x-y-6=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\x=7\end{cases}}\)

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

\(=x^2-2x\left(y+6\right)+\left(y+6\right)^2-\left(y+6\right)^2+6y^2+2y+45\)

\(=\left(x-\left(y+6\right)\right)^2-y^2-12y-36+6y^2+2y+45\)

\(=\left(x-y-6\right)^2+5y^2-10y+5+4=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)

Vậy \(A_{min}=4\)khi \(y=1\)và \(x=7\)

Ai giúp mik với ạ 😢😭😭😭😭😢😷

\(A=\left(x^2+y^2+36-2xy-12x+12y\right)+5y^2-10y+5+109\)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+109\ge109\)

\(A_{min}=109\) khi \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

(x^2+y^2-12y-12x+36)+(5y^2-10y+5)+4=(x-y-6)^2+5(y-1)^2+4>=4

GTNN A=4

khi y=1

x=7

A=\(\left(x-y\right)^2-2.6.\left(x-y\right)+36+5y^2+10y+5+4\)

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

Dấu bằng xảy ra khi y=1 và x=5

2B=\(2x^2+2y^2-2xy-2x+2y+2\)

=\(\left(x-y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)

=>B\(\ge\)0

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

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

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

Vậy giá trị nhỏ nhất của A = 4 khi :

\(\left\{{}\begin{matrix}y-1=0\\x-y-6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=7\end{matrix}\right.\)

A

Vậy nên giá trị nhỏ nhất của A = 4 khi :

A=x²- 2xy + 6y² - 12x + 2y + 45

A = (x² - 2xy + y² - 12x + 12y + 36) + (5y² - 10y + 5) + 4

= [(x - y)² - 12(x - y) + 6^2] + 5(y² - 2y + 1) + 4

= (x - y - 6)² + 5(y - 1)² + 4

Vì (x - y - 6)² >= 0 với mọi x, y

5(y² - 1) >= 0 với mọi y

=> A_min = 4 <=> y = 1, x = 7

\(A=\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(4y^2-12x+9\right)+35\)

\(=\left(x-y\right)^2+\left(y-1\right)^2+\left(2y-3\right)^2+35>=35\)

vậy gt A nhỏ nhất= 35 khi x=y, y=1, y=3/2