Question

Tìm GTNN của F=(mx+2y-2m)²+(x+y-2)²

Answer 1

- Với \(m=2\Rightarrow F=5\left(x+y-2\right)^2\ge0\)

\(F_{min}=0\) khi \(x+y=2\)

- Với \(m\ne2\)

\(\left\{{}\begin{matrix}\left(mx+2y-2m\right)^2\ge0\\\left(x+y-2\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow F_{min}=0\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}mx+2y=2m\\x+y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}mx+2y=2m\\2x+2y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-2\right)x=2\left(m-2\right)\\y=2-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)