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Câu hỏi: Giả sử $p,q$ là các số thực dương thỏa mãn ${{\log }_{16}}p={{\log }_{20}}q={{\log }_{25}}\left( p+q \right)$. Tìm giá trị của $\dfrac{p}{q}$.
A. $\dfrac{1}{2}\left( -1+\sqrt{5} \right)$.
B. $\dfrac{4}{5}$.
C. $\dfrac{1}{2}\left( 1+\sqrt{5} \right)$.
D. $\dfrac{8}{5}$.
A. $\dfrac{1}{2}\left( -1+\sqrt{5} \right)$.
B. $\dfrac{4}{5}$.
C. $\dfrac{1}{2}\left( 1+\sqrt{5} \right)$.
D. $\dfrac{8}{5}$.
Đặt $t={{\log }_{16}}p={{\log }_{20}}q={{\log }_{25}}\left( p+q \right)$
$\begin{aligned}
& \Rightarrow \left\{ \begin{aligned}
& p={{16}^{t}} \\
& q={{20}^{t}} \\
& p+q={{25}^{t}} \\
\end{aligned} \right. \\
& \Rightarrow {{16}^{t}}+{{20}^{t}}={{25}^{t}}\Leftrightarrow {{\left( \dfrac{4}{5} \right)}^{2t}}+{{\left( \dfrac{4}{5} \right)}^{t}}-1=0\Leftrightarrow {{\left( \dfrac{4}{5} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2} \\
\end{aligned}$
Mà $\dfrac{p}{q}=\dfrac{{{16}^{t}}}{{{20}^{t}}}={{\left( \dfrac{4}{5} \right)}^{t}}=\dfrac{1}{2}\left( -1+\sqrt{5} \right)$.
$\begin{aligned}
& \Rightarrow \left\{ \begin{aligned}
& p={{16}^{t}} \\
& q={{20}^{t}} \\
& p+q={{25}^{t}} \\
\end{aligned} \right. \\
& \Rightarrow {{16}^{t}}+{{20}^{t}}={{25}^{t}}\Leftrightarrow {{\left( \dfrac{4}{5} \right)}^{2t}}+{{\left( \dfrac{4}{5} \right)}^{t}}-1=0\Leftrightarrow {{\left( \dfrac{4}{5} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2} \\
\end{aligned}$
Mà $\dfrac{p}{q}=\dfrac{{{16}^{t}}}{{{20}^{t}}}={{\left( \dfrac{4}{5} \right)}^{t}}=\dfrac{1}{2}\left( -1+\sqrt{5} \right)$.
Đáp án A.