ta có : \(K=\dfrac{x^2}{x^2-5x+7}\Leftrightarrow\left(k-1\right)x^2-5kx+7k\)
vì phương trình này luôn có nghiệm \(\Rightarrow\Delta\ge0\)
\(\Leftrightarrow\left(5k\right)^2-4\left(k-1\right)7k\ge0\Leftrightarrow-3k^2+28k\ge0\)
\(\Leftrightarrow k\left(28-3k\right)\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}k\ge0\\28-3k\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}k\le0\\28-3k\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0\le k\le\dfrac{28}{3}\\k\in\varnothing\end{matrix}\right.\)
\(\Rightarrow0\le k\le\dfrac{28}{3}\)
\(\Rightarrow k_{max}=\dfrac{28}{3}\) khi \(x=\dfrac{-b}{2a}=\dfrac{5k}{2\left(k-1\right)}=\dfrac{5\left(\dfrac{28}{3}\right)}{2\left(\dfrac{28}{3}-1\right)}=\dfrac{14}{5}\)
\(\Rightarrow k_{min}=0\) khi \(x=\dfrac{-b}{2a}=\dfrac{5k}{2\left(k-1\right)}=\dfrac{5.0}{2\left(0-1\right)}=0\)vậy ...................................................................................................
