Question

Cho hệ phương trình

\(\left\{{}\begin{matrix}x+y=3m+2\\3x-2y=11-m\end{matrix}\right.\). Tìm m để hệ có nghiệm (x;y) thỏa mãn x²-y² đạt GTLN

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y=6m+4\\3x-2y=11-m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=5m+15\\3x-2y=11-m\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=m+3\\y=2m-1\end{matrix}\right.\)

\(A=x^2-y^2=\left(m+3\right)^2-\left(2m-1\right)^2\)

\(=-3m^2+10m+8=-3\left(m-\frac{5}{3}\right)^2+\frac{49}{3}\le\frac{49}{3}\)

\(A_{max}=\frac{49}{3}\) khi \(m=\frac{5}{3}\)