题目：已知实数 $m, n$ 满足 $\frac{e^{2025-2m}}{2} - m = \frac{e^{5-\ln 2}}{n} - \ln n - \ln(2e^{2020}) = 0$，则 $mn = \underline{\quad}$.
解：∵ $\frac{e^{2025-2m}}{2} - m = 0$
$\therefore e^{2025-2m} - 2m = 0$，
$\therefore e^{2025} = 2me^{2m}$，
$\therefore \ln e^{2025} = \ln 2m + \ln e^{2m}$
$\therefore 2025 = \ln 2m + 2m$
$\therefore e^{\ln 2m} + \ln 2m = 2025$.
又 $\because \frac{e^{5-\ln 2}}{n} - \ln n - \ln(2e^{2020}) = 0$
$\therefore \frac{e^{5-\ln 2}}{e^{\ln n}} - \ln n - \ln 2 - 2020 = 0$，
$\therefore e^{5-\ln 2n} + 5 - \ln 2n = 2025$，
设 $f(x) = x + e^x$，则 $f'(x) = 1 + e^x > 0$
$\therefore$ 函数 $f(x)$ 在 $\mathbb{R}$ 上单调递增，
$\therefore f(\ln 2m) = f(5 - \ln 2n) = 2025$，
$\therefore \ln 2m = 5 - \ln 2n$
$\therefore \ln 4mn = 5$，
$\therefore mn = \frac{e^5}{4}$.