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Câu hỏi: Cho 2 số thực dương thỏa mãn: ${{\log }_{9}}a={{\log }_{15}}b={{\log }_{25}}(a+b)$. Tính $\dfrac{a}{b}$.
A. $\dfrac{1}{2}$.
B. $\dfrac{-1-\sqrt{5}}{2}$.
C. $\dfrac{1+\sqrt{5}}{2}$.
D. $\dfrac{-1+\sqrt{5}}{2}$.
A. $\dfrac{1}{2}$.
B. $\dfrac{-1-\sqrt{5}}{2}$.
C. $\dfrac{1+\sqrt{5}}{2}$.
D. $\dfrac{-1+\sqrt{5}}{2}$.
Đặt ${{\log }_{9}}a={{\log }_{15}}b={{\log }_{25}}(a+b)=t$
Suy ra $a={{9}^{t}},b={{15}^{t}},a+b={{25}^{t}}$.
Ta có phương trình: ${{9}^{t}}+{{15}^{t}}={{25}^{t}}\Leftrightarrow {{\left( \dfrac{3}{5} \right)}^{2t}}+{{\left( \dfrac{3}{5} \right)}^{t}}-1=0\Leftrightarrow \left[ \begin{aligned}
& {{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{-1-\sqrt{5}}{2}(l) \\
& {{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2}(tm) \\
\end{aligned} \right.$
Vậy $\dfrac{a}{b}={{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2}.$
Suy ra $a={{9}^{t}},b={{15}^{t}},a+b={{25}^{t}}$.
Ta có phương trình: ${{9}^{t}}+{{15}^{t}}={{25}^{t}}\Leftrightarrow {{\left( \dfrac{3}{5} \right)}^{2t}}+{{\left( \dfrac{3}{5} \right)}^{t}}-1=0\Leftrightarrow \left[ \begin{aligned}
& {{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{-1-\sqrt{5}}{2}(l) \\
& {{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2}(tm) \\
\end{aligned} \right.$
Vậy $\dfrac{a}{b}={{\left( \dfrac{3}{5} \right)}^{t}}=\dfrac{-1+\sqrt{5}}{2}.$
Đáp án D.