Cho x;y > 0 thỏa mãn log9(x)=log6(y)=log(x+y). Tính x/y

* Hướng dẫn giải

Đặt ${\log }_9\left(x\right)={\log }_6\left(y\right)=\log \left(x+y\right)$ = t

$\Rightarrow x=9^t;y=6^t;x+y=4^t$

Khi đó

$$\begin{gathered} 9^t+6^t=4^t \\ \iff {\left(\frac{3}{2}\right)}^{2t}+{\left(\frac{3}{2}\right)}^t-1=0 \\ \iff \left\{\begin{array}{l}{\left(\frac{3}{2}\right)}^t=\frac{\sqrt{5}-1}{2}\\ {\left(\frac{3}{2}\right)}^t=\frac{-\sqrt{5}-1}{2}<0\end{array}\right. \end{gathered}$$

Hơn nữa $\frac{x}{y}=\frac{9^t}{6^t}={\left(\frac{3}{2}\right)}^t$ = $\frac{\sqrt{5}-1}{2}$

Đáp án C