Cho \(a > 0;\,b > 0\) thỏa điều kiện: \({\log _9}a = {\log _{12}}b = {\log _{16}}\left( {a + b} \right)\).

* Hướng dẫn giải

Đặt:\({\log _9}a = {\log _{12}}b = {\log _{16}}\left( {a + b} \right) = t \Rightarrow \left\{ \begin{array}{l}
a = {9^t}\\
b = {12^t}\\
a + b = {16^t}
\end{array} \right.\)

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

\(\begin{array}{l}
 = {\log _4}2 + {\log _8}\sqrt 2  + {\log _{16}}\sqrt[3]{2} + ... + {\log _{{2^{2018}}}}\sqrt[{2017}]{2} = \frac{1}{2} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{2017.2018}}\\
 = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ... + \left( {\frac{1}{{2017}} - \frac{1}{{2018}}} \right) = 1 - \frac{1}{{2018}} = \frac{{2017}}{{2018}}
\end{array}\)