Cho số thực dương x, y thỏa mãn

B

- Hướng dẫn giải

Đáp án B.

Đặt

$$\begin{gathered} {\log }_{6}x={\log }_{9}y={\log }_4\left(2x+2y\right)=t \\ \Rightarrow \left\{\begin{array}{l}x=6^t\\ y=6^t\\ 2x+2y=4^t\end{array}\right.\Rightarrow 2\left(6^t+9^t\right)=4^t \\ \iff 2\left[{\left(\frac{6}{9}\right)}^t+1\right]={\left(\frac{4}{9}\right)}^t \\ \iff 2\left[{\left(\frac{2}{3}\right)}^t+1\right]={\left(\frac{2}{3}\right)}^{2t}. \end{gathered}$$

Đặt $u={\left(\frac{2}{3}\right)}^t=\frac{x}{y}>0$ ta có:

$$\begin{gathered} 2\left(u+1\right)=u^2 \\ \Rightarrow u=1+\sqrt{3}=\frac{2}{\sqrt{3}-1}. \end{gathered}$$