Google
Câu hỏi: Cho các số thực dương $a,b$ thỏa mãn ${{\log }_{9}}a={{\log }_{12}}b={{\log }_{16}}(a+b)$. Tính tỉ số $\dfrac{a}{b}$.
A. $\dfrac{a}{b}=\dfrac{1-\sqrt{5}}{2}$.
B. $\dfrac{a}{b}=\dfrac{-1-\sqrt{5}}{2}$
C. $\dfrac{a}{b}=\dfrac{-1+\sqrt{5}}{2}$.
D. $\dfrac{a}{b}=\dfrac{1+\sqrt{5}}{2}$.
Đặt ${{\log }_{9}}a={{\log }_{12}}b={{\log }_{16}}(a+b)=t$
$\Rightarrow \left\{ \begin{matrix}
a={{9}^{t}} \\
b={{12}^{t}} \\
a+b={{16}^{t}} \\
\end{matrix}\Rightarrow {{16}^{t}}={{12}^{t}}+{{9}^{t}}\Rightarrow 1-{{\left( \dfrac{3}{4} \right)}^{t}}-{{\left( \dfrac{3}{4} \right)}^{2t}}=0 \right.$$\Rightarrow {{\left( \dfrac{3}{4} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2}=\dfrac{a}{b}$.
A. $\dfrac{a}{b}=\dfrac{1-\sqrt{5}}{2}$.
B. $\dfrac{a}{b}=\dfrac{-1-\sqrt{5}}{2}$
C. $\dfrac{a}{b}=\dfrac{-1+\sqrt{5}}{2}$.
D. $\dfrac{a}{b}=\dfrac{1+\sqrt{5}}{2}$.
:
Đặt ${{\log }_{9}}a={{\log }_{12}}b={{\log }_{16}}(a+b)=t$
$\Rightarrow \left\{ \begin{matrix}
a={{9}^{t}} \\
b={{12}^{t}} \\
a+b={{16}^{t}} \\
\end{matrix}\Rightarrow {{16}^{t}}={{12}^{t}}+{{9}^{t}}\Rightarrow 1-{{\left( \dfrac{3}{4} \right)}^{t}}-{{\left( \dfrac{3}{4} \right)}^{2t}}=0 \right.$$\Rightarrow {{\left( \dfrac{3}{4} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2}=\dfrac{a}{b}$.
Đáp án C.