Подставим \( n = -\frac{5}{12} \) в выражение:
\[ \left(-\frac{5}{12} + 6\right)^2 + \left(2 \cdot \left(-\frac{5}{12}\right)\right) \left(2 + \left(-\frac{5}{12}\right)\right) \]Упростим выражение:
\[ \left(-\frac{5}{12} + \frac{72}{12}\right)^2 + \left(-\frac{10}{12}\right) \left(\frac{24}{12} - \frac{5}{12}\right) \] \[ \left(\frac{67}{12}\right)^2 + \left(-\frac{5}{6}\right) \left(\frac{19}{12}\right) \] \[ \frac{4489}{144} - \frac{95}{72} \] \[ \frac{4489}{144} - \frac{190}{144} \] \[ \frac{4299}{144} = \frac{1433}{48} = 29\frac{41}{48} \]Ответ: \(\frac{1433}{48}\) или 29\(\frac{41}{48}\)