A=2x^2+9y^2-6xy-6x-12y+2024 
A = (x^2 -6xy +9y^2) + 4(x -3y) + x^2 - 10x + 2024
A = (x -3y)^2 +4(x -3y) + 4 + x^2 -10x +25 + 1995
A = (x -3y +2)^2 + (x -5)^2 + 1995 \geq 1995
Min A = 1995 
 x - 5 = 0 => x = 5
Và x - 3y + 2 = 0 hay 5 -3y +2 = 0 => -3y = -7 => y = 7/3 

\(K\)\(nha!~!\)

anh ko biết nha em yêu của anh

\(A=2x^2+9y^2-6xy-6x-12y+2046\)

\(=\left[\left(x^2-6xy+9y^2\right)+\left(4x-12y\right)+4\right]-4+\left(x^2-10x+25\right)-25+2046\)

\(=\left[\left(x-3y\right)^2+4\left(x-3y\right)+4\right]+\left(x-5\right)^2-4-25+2046\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+2017\ge2017\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-3y+2=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=\frac{7}{3}\\x=5\end{cases}}}\)

Vậy \(A_{min}=2017\) tại \(x=5;y=\frac{7}{3}\)

\(A=2x^2+9y^2-6xy-6x-12y+2015\)

\(A=\left(x^2-6xy+9y^2\right)+x^2-6x-12y+2015\)

\(A=\left(x-3y\right)^2+4.\left(x-3y\right)-10x+x^2+2015\)

\(A=\left(x-3y\right)^2+4.\left(x-3y\right)+4+\left(x^2-10x+25\right)+1986\)

\(A=\left(x-3y+2\right)^2+\left(x-5\right)^2+1986\)

Vì \(\left(x-3y+2\right)^2\ge0;\left(x-5\right)^2\ge0\)

\(\Rightarrow A\ge1986\)

Dấu '=' xảy ra khi:

\(\Rightarrow\hept{\begin{cases}x-3y+2=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{7}{3}\\x=5\end{cases}}}\)

Vậy A_min= 1986 khi x = 5, y = 7/3

Chúc bạn học tốt!!!