\(9x^2+5y^2-6xy-6x-6y+20\)

\(=9x^2+y^2+1-6x+2y-6xy+4y^2-8y+4+15\)

\(=\left(3x-y-1\right)^2+4\left(y-1\right)^2+15\ge15\)

Dấu \(=\)khi \(\hept{\begin{cases}3x-y-1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=1\end{cases}}\).

A = 9x² - 6xy + 5y² + 1 = (3x)² + 2.3y + y² + (2y)² + 1 = ( 3x + y)² + ( 2y )² +1 
mà ( 3x + y)² > 0 và ( 2y )² > 0 

=> ( 3x + y )² + (2y)² + 1 > 0

Vậy gtnn của A là 1 

\(a,A=2x^2+9y^2-6xy-6x-12y+2049\)

\(=x^2-6xy+9y^2+x^2-10x+25+4x-12y+2024\)

\(=\left(x-3y\right)^2+\left(x-5\right)^2+4\left(x-3y\right)+2024\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-5\right)^2+2020\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+2020\)

\(A_{min}=2020\Leftrightarrow\hept{\begin{cases}\left(x-3y+2\right)^2=0\\\left(x-5\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-3y+2=0\\x-5=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x-3y+2=0\\x=5\end{cases}\Rightarrow5-3y+2=0}\)

\(\Rightarrow3y=7\Leftrightarrow y=\frac{7}{3}\)

Vậy \(A_{min}=2020\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}\)

b tương tự nhé

\(C=1-6y-5y^2-12xy-9x^2\)

\(\Rightarrow C=-4y^2-12xy-9x^2-y^2-6y+1\)

\(\Rightarrow C=-\left(4y^2+12xy+9x^2\right)-\left(y^2+6y+9\right)+1+9\)

\(\Rightarrow C=-\left(2y-3x\right)^2-\left(y+3\right)^2+10\)

mà \(\left\{{}\begin{matrix}-\left(2y-3x\right)^2\le0,\forall x;y\\-\left(y+3\right)^2\le0,\forall y\end{matrix}\right.\)

\(\Rightarrow C=-\left(2y-3x\right)^2-\left(y+3\right)^2+10\le10\)

\(\Rightarrow GTLN\left(C\right)=10\left(tạix=-2;y=-3\right)\)