Упростим выражение:
\(\frac{a^2+9}{2a+8} \cdot \frac{4a+16}{a^2+6a+9} = \frac{a^2+9}{2(a+4)} \cdot \frac{4(a+4)}{(a+3)^2} = \frac{a^2+9}{1} \cdot \frac{2}{(a+3)^2} = \frac{2(a^2+9)}{(a+3)^2}\)
Теперь подставим \(a = 1,8\):
\(a+3 = 1,8+3 = 4,8\)
\((a+3)^2 = (4,8)^2 = 23,04\)
\(a^2 = (1,8)^2 = 3,24\)
\(a^2+9 = 3,24+9 = 12,24\)
\(2(a^2+9) = 2(12,24) = 24,48\)
\(\frac{24,48}{23,04} = \frac{2448}{2304} = \frac{102}{96} = \frac{51}{48} = \frac{17}{16}\)
Ответ: \(\frac{17}{16}\).