Đáp án:
$\dfrac{p}{q} = \dfrac{-1 + \sqrt5}{2}$
Giải thích các bước giải:
Đặt $\log_9p = \log_{12}q = \log_{16}(p+q) = t$
$\Rightarrow \begin{cases}9^t = p\\12^t = q\\16^t = p + q\end{cases}$
Ta được:
$\quad 9^t + 12^t = 16^t$
$\Leftrightarrow \dfrac{9^t}{16^t} + \dfrac{12^t}{16^t} - 1 =0$
$\Leftrightarrow \left(\dfrac34\right)^{2t} + \left(\dfrac34\right)^t - 1 = 0$
$\Leftrightarrow \left[\begin{array}{l}\left(\dfrac34\right)^t = \dfrac{-1-\sqrt5}{2}\quad (vn)\\\left(\dfrac34\right)^t = \dfrac{-1 + \sqrt5}{2}\end{array}\right.$
$\Leftrightarrow \left(\dfrac34\right)^t = \dfrac{-1 + \sqrt5}{2}$
$\Leftrightarrow \left(\dfrac{9}{12}\right)^t = \dfrac{-1 + \sqrt5}{2}$
$\Leftrightarrow \dfrac{9^t}{12^t} = \dfrac{-1 + \sqrt5}{2}$
$\Leftrightarrow \dfrac{p}{q} = \dfrac{-1 + \sqrt5}{2}$
