\(A=\dfrac{x^2+mx+n}{x^2+2x+4}\)
\(\Leftrightarrow Ax^2+2Ax+4A=x^2+mx+n\)
\(\Leftrightarrow\left(A-1\right)x^2+\left(2A-m\right)x+\left(4A-n\right)=0\left(1\right)\)
A có cực trị khi (1) có nghiệm
\(\Leftrightarrow\Delta=\left(4A^2-4Am+m^2\right)-4\left[4A^2-A\left(n+4\right)+n\right]\ge0\)
\(\Leftrightarrow-12A^2-4A\left(m-n-4\right)+m^2-4n\ge0\) (1)
Mặt khác, theo gt, ta có: \(\left\{{}\begin{matrix}A\ge\dfrac{1}{3}\\A\le3\end{matrix}\right.\)
\(\Rightarrow\left(3A-1\right)\left(3-A\right)\ge0\)
\(\Leftrightarrow-3A^2+10A-3\ge0\)
\(\Leftrightarrow-12A^2+40A-12\ge0\) (2)
Từ (1) và (2) suy ra \(\left\{{}\begin{matrix}m-n-4=-10\\m^2-4n=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m+6=n\\m^2-4\left(m+6\right)=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n=m+6\\\left(m-6\right)\left(m+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}n=12\\n=4\end{matrix}\right.\\\left[{}\begin{matrix}m=6\\m=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(m;n\right)=\left(6;12\right);\left(-2;4\right)\)
