Cho các số thực dương x,y thỏa mãn...

C

- Hướng dẫn giải

Đáp án C

Đặt

$\begin{array}{l}{\log }_{9}x={\log }_{12}y\\ ={\log }_{16}\left(x+y\right)=t\\ \Rightarrow \left\{\begin{array}{l}x=9^t\\ y={12}^t\\ x+y={16}^t\end{array}\right.\\ \Rightarrow 9^t+{12}^t={16}^t\end{array}$

$\iff {\left(3^t\right)}^2+3^t{.4}^t-{\left(4^t\right)}^2=0\left(*\right)$

Chia cả hai vế của phương trình (*) cho ${\left(4^t\right)}^2$ ta được:

$\begin{array}{l}{\left(\frac{3^t}{4^t}\right)}^2+\frac{3^t}{4^t}-1=0\\ \iff \left[\begin{array}{l}\frac{3^t}{4^t}=\frac{\sqrt{5}-1}{2}\\ \frac{3^t}{4^t}=\frac{-\sqrt{5}-1}{2}\left(L\right)\end{array}\right.\\ \Rightarrow \frac{x}{y}=\frac{3^t}{4^t}=\frac{\sqrt{5}-1}{2}\end{array}$

Ta có:

$S={\log }_4\frac{x\left(1+\sqrt{5}\right)}{y}$$+{\log }_8\sqrt{\frac{x\left(1+\sqrt{5}\right)}{y}}+{\log }_{16}\sqrt[3]{\frac{x\left(1+\sqrt{5}\right)}{y}}$$+\mathrm{...}+{\log }_{2^{2018}}\sqrt[2017]{\frac{x\left(1+\sqrt{5}\right)}{y}}$

$={\log }_{2^2}\frac{x\left(1+\sqrt{5}\right)}{y}+{\log }_{2^3}{\left[\frac{x\left(1+\sqrt{5}\right)}{y}\right]}^{\frac{1}{2}}$$+{\log }_{2^4}{\left[\frac{x\left(1+\sqrt{5}\right)}{y}\right]}^{\frac{1}{3}}+\mathrm{...}+$${\log }_{2^{2018}}{\left[\frac{x\left(1+\sqrt{5}\right)}{y}\right]}^{\frac{1}{2017}}$

$=\frac{1}{1.2}{\log }_2\frac{x\left(1+\sqrt{5}\right)}{y}$$+\frac{1}{2.3}{\log }_2\frac{x\left(1+\sqrt{5}\right)}{y}$$+\frac{1}{3.4}{\log }_2\frac{x\left(1+\sqrt{5}\right)}{y}+\mathrm{...}+$$\frac{1}{2017.2018}{\log }_2\frac{x\left(1+\sqrt{5}\right)}{y}$

$=( \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)$$+\left(\frac{1}{3}-\frac{1}{4}\right)+\mathrm{...}+\left(\frac{1}{2017}-\frac{1}{2018}\right) )$$.{\log }_2\frac{x\left(1+\sqrt{5}\right)}{y}$

$=\left(1-\frac{1}{2018}\right).{\log }_2\frac{x\left(1+\sqrt{5}\right)}{y}$$=\frac{2017}{2018}.{\log }_2\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}{2}$$=\frac{2017}{2018}$