\(A=x^2-2xy+6y^2-12x+3y+45\)
\(A=x^2-2x\left(y+6\right)+6y^2+3y+45\)
\(A=x^2-2x\left(y+6\right)+y^2+2.y.6+36+5y^2-9y+9\)
\(A=x^2-2x\left(y+6\right)+\left(y+6\right)^2+5\left(y^2-2.y.\frac{9}{10}+\frac{81}{100}\right)-\frac{81}{20}+9\)
\(A=\left(x-y-6\right)^2+5\left(y-\frac{9}{10}\right)^2-\frac{99}{20}\)
Ta thấy: \(\left(x-y-6\right)^2\ge0;5\left(y-\frac{9}{10}\right)^2\ge0\forall x;y\)
\(\Rightarrow A\ge-\frac{99}{20}.\)Vậy \(Min_A=-\frac{99}{20}.\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y-6=0\\y-\frac{9}{10}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=6\\y=\frac{9}{10}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{69}{10}\\y=\frac{9}{10}\end{cases}}.\)
Xin lỗi, \(Min_A=\frac{99}{20}\)nha bạn, vì \(-\frac{81}{20}+9=-\left(\frac{81}{20}-9\right)=-\left(-\frac{99}{20}\right)=\frac{99}{20}.\)
