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