\(\left\{{}\begin{matrix}x+y=3m+2\\3x-2y=11-m\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}3x+3y=9m+6\\3x-2y=11-m\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}5y=10m-5\\x+y=3m+2\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}y=2m-1\\x=3m+2-y\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}y=2m-1\\x=m+3\end{matrix}\right.\)
\(\rightarrow x^2-y^2=\left(m+3\right)^2-\left(2m-1\right)^2\)
\(=-3m^2+10m+8\)
\(=-\left(3m^2-10m-8\right)\)
\(=-3\left(m^2-2.\frac{5}{3}m+\frac{25}{9}\right)-\frac{1}{3}\)
\(=-3\left(m-\frac{5}{3}\right)^2-\frac{1}{3}\le-\frac{1}{3}\forall m\)
Dấu = xảy ra khi \(m=\frac{5}{3}\)
Vậy \(m=\frac{5}{3}\) thì biểu thức đạt giá trị lớn nhất là \(-\frac{1}{3}\)