Example 5. Given the function $y=\frac{5 x^{2}-2 k x+50}{4 x^{2}-10 x+25}$ has a minimum value of 1, find the real number $\mathrm{k}$.
Misunderstanding: $\because \frac{5 x^{2}-2 kx+50}{4 x^{2}-10 x+25} \geqslant 1$, and $4 x^{2}-10 x+25=4\left(x-\frac{5}{4}\right)^{2}+\frac{75}{4}>0$, $\therefore 5 x^{2}-2 kx+50 \geqslant 4 x^{2}-10 x+25$, i.e., $x^{2}-2(k-5) x+25 \geqslant 0, \quad \therefore \Delta \leqslant 0$, i.e., $(k-5)^{2}-25 \leqslant 0$, solving it yields $0 \leqslant k \leqslant 10$.

Analysis: This problem should be understood as finding the set of real numbers $k$ that make the function $y$ have a minimum value of 1. Therefore, at the beginning, one should focus on the necessary and sufficient condition for $y$ to have a minimum value of 1, which is $\frac{5 x^{2}-2 kx+50}{4 x^{2}-10 x+25} \geqslant 1$ (the equality must hold) $\Leftrightarrow x^{2}-2(k-5) x +25 \geqslant 0$ (the equality must hold) $\Leftrightarrow \Delta=0$, i.e., $(k-5)^{2}=25 \Leftrightarrow|k-5|=5 \Leftrightarrow k=0$ or $k=10$.

The flaw in the solution by Qi Sha is that it incorrectly judges the necessary and sufficient condition for the non-strict inequality $x^{2}-2(k-5) x+25 \geqslant 0$ to hold as $\Delta \leqslant 0$. In fact, when $\Delta=0$, it does not guarantee that $y$ has a minimum value of 1.