- Với \(a=2\) đặt \(2x+y=t\Rightarrow F=\left(t-2\right)^2+\left(2t-1\right)^2\)
\(F=5t^2-8t+5=5\left(t-\frac{4}{5}\right)^2+\frac{9}{5}\ge\frac{9}{5}\)
\(F_{min}=\frac{9}{5}\) khi \(t=\frac{4}{5}\Leftrightarrow2x+y=\frac{4}{5}\)
- Với \(a\ne2\):
Do \(\left\{{}\begin{matrix}\left(2x+y-2\right)^2\ge0\\\left(4x+ay-1\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow F\ge0\Rightarrow F_{min}=0\) khi:
\(\left\{{}\begin{matrix}2x+y-2=0\\4x+ay-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=4\\4x+ay=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{2a+1}{2\left(a-2\right)}\\y=-\frac{3}{a-2}\end{matrix}\right.\)