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Câu hỏi: Cho $a,b$ là các số thực dương thỏa mãn $\log _{27} a+\log _{9} b^{2}=5$ và $\log _{9} a^{2}+\log _{27} b=7$. Giá trị của $a \cdot b$ bằng
A. ${{3}^{12}}$.
B. $3^{16}$.
C. $3^{18}$.
D. $3^{9}$.
A. ${{3}^{12}}$.
B. $3^{16}$.
C. $3^{18}$.
D. $3^{9}$.
+) $\left\{ \begin{aligned}
& {{\log }_{27}}a+{{\log }_{9}}{{b}^{2}}=5 \\
& {{\log }_{9}}{{a}^{2}}+{{\log }_{27}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{3}{{\log }_{3}}a+\dfrac{2}{2}{{\log }_{3}}b=5 \\
& \dfrac{2}{2}{{\log }_{3}}a+\dfrac{1}{3}{{\log }_{3}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{3}}a+3{{\log }_{3}}b=15 \\
& 3{{\log }_{3}}a+{{\log }_{3}}b=21 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{3}}a=6 \\
& {{\log }_{3}}b=3 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a={{3}^{6}} \\
& b={{3}^{3}} \\
\end{aligned} \right.\Rightarrow a.b={{3}^{6}}{{.3}^{3}}={{3}^{9}}$.
& {{\log }_{27}}a+{{\log }_{9}}{{b}^{2}}=5 \\
& {{\log }_{9}}{{a}^{2}}+{{\log }_{27}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{3}{{\log }_{3}}a+\dfrac{2}{2}{{\log }_{3}}b=5 \\
& \dfrac{2}{2}{{\log }_{3}}a+\dfrac{1}{3}{{\log }_{3}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{3}}a+3{{\log }_{3}}b=15 \\
& 3{{\log }_{3}}a+{{\log }_{3}}b=21 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{3}}a=6 \\
& {{\log }_{3}}b=3 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a={{3}^{6}} \\
& b={{3}^{3}} \\
\end{aligned} \right.\Rightarrow a.b={{3}^{6}}{{.3}^{3}}={{3}^{9}}$.
Đáp án D.