\(\begin{array}{l}\begin{array}{*{20}{l}}{OA = dm}\\{AB = {\rm{ 9}}m}\end{array}\\AM{\rm{ }} = 4,5m\end{array}\)

\(\begin{array}{l}\tan \angle MOB = \tan ({\alpha _1} - {\alpha _2}) = \frac{{\tan {\alpha _1} - \tan {\alpha _2}}}{{1 + \tan {\alpha _1}\tan {\alpha _2}}}\\ = \frac{{\frac{9}{d} - \frac{{4,5}}{d}}}{{1 + \frac{9}{d}.\frac{{4,5}}{d}}} = \frac{{4,5}}{{d + \frac{{81}}{{2d}}}}\end{array}\)

+ Theo BĐT Cosi, ta có:

\(d + \frac{{81}}{{2d}} \ge 2\sqrt {\frac{{81}}{2}} = 9\sqrt 2 \)

Dấu “=” xảy ra khi: \(d = \frac{{81}}{2}d \to d = 4,5\sqrt 2 m\)

Do đó: \(OM = \sqrt {{{(4,5\sqrt 2 )}^2} + 4,{5^2}} = 4,5\sqrt 3 m\)

+ Ta có:

\(\begin{array}{l}{L_A} - {L_M} = 10\log \frac{{{I_A}}}{{{I_M}}}\\ \leftrightarrow 60 - 80 = - 20 = 10\log \frac{{{I_A}}}{{{I_M}}}\\ \to \frac{{{I_A}}}{{{I_M}}} = 0,01\end{array}\)

Mặt khác:

\(\begin{array}{l}\left\{ \begin{array}{l}{I_A} = \frac{{2P}}{{4\pi R_A^2}}\\{I_M} = \frac{{(x + 2)P}}{{4\pi R_M^2}}\end{array} \right.\\ \to \frac{{{I_A}}}{{{I_M}}} = \frac{2}{{x + 2}}\frac{{R_M^2}}{{R_A^2}} = \frac{2}{{x + 2}}\frac{{{{\left( {4,5\sqrt 3 } \right)}^2}}}{{{{\left( {4,5\sqrt 2 } \right)}^2}}} = 0,01\\ \to x = 298\end{array}\)