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Câu hỏi: Với hai số thực $a,b$ tuỳ ý thoả mãn $\dfrac{{{\log }_{3}}5{{\log }_{5}}a}{1+{{\log }_{3}}2}-{{\log }_{6}}b=2$. Khẳng định nào dưới đây đúng
A. $a=36b$.
B. $2\text{a}+3b=0$
C. $a=b{{\log }_{6}}2$
D. $a=b{{\log }_{6}}3$
A. $a=36b$.
B. $2\text{a}+3b=0$
C. $a=b{{\log }_{6}}2$
D. $a=b{{\log }_{6}}3$
$\begin{aligned}
& \dfrac{{{\log }_{3}}5{{\log }_{5}}a}{1+{{\log }_{3}}2}-{{\log }_{6}}b=2\Leftrightarrow \dfrac{{{\log }_{3}}a}{{{\log }_{3}}6}-{{\log }_{6}}b=2\Leftrightarrow {{\log }_{3}}a-{{\log }_{3}}6{{\log }_{6}}b=2{{\log }_{3}}6 \\
& \Leftrightarrow {{\log }_{3}}a-{{\log }_{3}}b={{\log }_{3}}36\Leftrightarrow {{\log }_{3}}\dfrac{a}{b}={{\log }_{3}}36\Leftrightarrow \dfrac{a}{b}=36\Leftrightarrow a=36b \\
\end{aligned}$
& \dfrac{{{\log }_{3}}5{{\log }_{5}}a}{1+{{\log }_{3}}2}-{{\log }_{6}}b=2\Leftrightarrow \dfrac{{{\log }_{3}}a}{{{\log }_{3}}6}-{{\log }_{6}}b=2\Leftrightarrow {{\log }_{3}}a-{{\log }_{3}}6{{\log }_{6}}b=2{{\log }_{3}}6 \\
& \Leftrightarrow {{\log }_{3}}a-{{\log }_{3}}b={{\log }_{3}}36\Leftrightarrow {{\log }_{3}}\dfrac{a}{b}={{\log }_{3}}36\Leftrightarrow \dfrac{a}{b}=36\Leftrightarrow a=36b \\
\end{aligned}$
Đáp án A.