\(A=3x^2+2\left(y-5\right)x+2y^2-10y+2030\)
\(\Leftrightarrow3x^2+2\left(y-5\right)x+2y^2-10y+2030-A=0\)
Để tồn tại x, y thỏa mãn, ta phải có:
\(\Delta'=\left(y-5\right)^2-3\left(2y^2-10y+2030-A\right)\ge0\)
\(\Leftrightarrow-5y^2+20y-6065+3A\ge0\)
\(\Leftrightarrow3A\ge5y^2-20y+6065=5\left(y^2-4y+4\right)+6045\)
\(\Leftrightarrow3A\ge5\left(y-2\right)^2+6045\Rightarrow A\ge\dfrac{5}{3}\left(y-2\right)^2+2015\ge2015\)
\(\Rightarrow A_{min}=2015\) khi \(y=2\Rightarrow x=1\)
Làm theo kiểu lớp 8 thì như sau:
\(A=2y^2+2y\left(x-5\right)+3x^2-10x+2030\)
\(A=2\left(y^2+2y.\dfrac{\left(x-5\right)}{2}+\left(\dfrac{x-5}{2}\right)^2\right)+\dfrac{5}{2}\left(x^2-2x+1\right)+2025\)
\(A=2\left(y+\dfrac{x-5}{2}\right)^2+\dfrac{5}{2}\left(x-1\right)^2+2025\ge2025\)
\(\Rightarrow A_{min}=2025\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x-1=0\\y+\dfrac{x-5}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
