Toán 12 Số thực dương a, b thỏa mãn \({\log _9}a = {\log _{12}}b = {\log _{16}}\left( {a + b} \right)\)

Số thực dương a, b thỏa mãn \({\log _9}a = {\log _{12}}b = {\log _{16}}\left( {a + b} \right)\). Mệnh đề nào dưới đây đúng?
A. \(\frac{a}{b} \in \left( {\frac{2}{3};1} \right)\)
B. \(\frac{a}{b} \in \left( {0;\frac{2}{3}} \right)\)
C. \(\frac{a}{b} \in \left( {9;12} \right)\)
D. \(\frac{a}{b} \in \left( {9;16} \right)\)

Đặt \(t = {\log _9}a = {\log _{12}}b = {\log _{16}}\left( {a + b} \right) \Rightarrow \left\{ \begin{array}{l}a = {9^t}\\b = {12^t}\\a + b = {16^t}\left( * \right)\end{array} \right. \Rightarrow \frac{a}{b} = {\left( {\frac{3}{4}} \right)^t}\)
\(\left( * \right) \Leftrightarrow {9^t} + {12^t} = {16^t} \Leftrightarrow {\left( {\frac{3}{4}} \right)^t} + 1 = {\left( {\frac{4}{3}} \right)^t} \Leftrightarrow {\left( {\frac{3}{4}} \right)^{2t}} + {\left( {\frac{3}{4}} \right)^t} - 1 = 0 \Leftrightarrow \left[ \begin{array}{l}{\left( {\frac{3}{4}} \right)^t} = \frac{{ - 1 + \sqrt 5 }}{2}\\{\left( {\frac{3}{4}} \right)^t} = \frac{{ - 1 - \sqrt 5 }}{2}\end{array} \right.\)
\( \Rightarrow {\left( {\frac{3}{4}} \right)^t} = \frac{{ - 1 + \sqrt 5 }}{2} \Leftrightarrow \frac{a}{b} = \frac{{ - 1 + \sqrt 5 }}{2} \Rightarrow \frac{a}{b} \in \left( {0;\frac{2}{3}} \right).\)