Bài toán tương đương: tìm m để pt \(x^2-\left(2m-3\right)x-m^2+3m=0\) có 2 nghiệm pb thỏa mãn \(1< x_1< x_2< 6\)
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(2m-3\right)^2-4\left(-m^2+3m\right)>0\\f\left(1\right)=1-\left(2m-3\right)-m^2+3m>0\\f\left(6\right)=36-6\left(2m-3\right)-m^2+3m>0\\1< \frac{x_1+x_2}{2}< 6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8m^2-24m+9>0\\-m^2+m+4>0\\-m^2-9m+54>0\\2< 2m-3< 12\end{matrix}\right.\) \(\Rightarrow\frac{6+3\sqrt{2}}{4}< m< \frac{1+\sqrt{17}}{2}\)
