421 ** Find the maximum value of the function $y=\frac{(x-1)^{5}}{(10 x-6)^{9}}$ for $x>1$.


Parse $y=\frac{1}{2^{5}} \cdot \frac{(2 x-2)^{5}}{(10 x-6)^{9}}=\frac{1}{2^{5}} \cdot\left(\frac{2 x-2}{10 x-6}\right)^{5} \cdot\left(\frac{1}{10 x-6}\right)^{4}$,
$$
\leqslant \frac{1}{2^{5}} \cdot \frac{\left[5 \cdot \frac{2 x-2}{10 x-6}+\frac{4}{10 x-6}\right]^{9}}{9^{9}}=\frac{1}{2^{5} \cdot 9^{9}},
$$

Therefore, $y_{\text {max }}=\frac{1}{2^{5} \cdot 9^{9}}$, with equality when $2 x-2=1$ or $x=\frac{3}{2}$.
Comment: This problem demonstrates the use of the method of undetermined coefficients and the AM-GM inequality to find the maximum value. Initially, we set
$$
y=\frac{1}{\lambda^{5}} \cdot \frac{(\lambda x-\lambda)^{5}}{(10 x-6)^{9}} \leqslant \frac{1}{\lambda^{5}} \cdot \frac{\left[5 \cdot \frac{\lambda x-\lambda}{10 x-6}+4 \cdot \frac{1}{10 x-6}\right]^{9}}{9^{9}},
$$

to obtain a constant. Let $5(\lambda x-\lambda)+4=k(10 x-6)$, so we just need to take $\lambda=2, k=1$, thus determining the undetermined coefficients.