Cho $a>0,b>0$, nếu viết $\displaystyle {{\log }_{3}}{{\left( \sqrt[5]{{{a}^{3}}b} \right)}^{\frac{2}{3}}}=\frac{x}{5}{{\log }_{3}}a+\frac{y}{15}{{\log }_{3}}b$ thì $x+y$ bằng A. $3.$ B. $5.$ C. $2.$ D. $4.$
Đáp án đúng: D Ta có: $\displaystyle {{\log }_{3}}{{\left( \sqrt[5]{{{a}^{3}}b} \right)}^{\frac{2}{3}}}={{\log }_{3}}{{({{a}^{3}}b)}^{\frac{2}{15}}}=\frac{2}{5}{{\log }_{3}}a+\frac{2}{15}{{\log }_{3}}b\Rightarrow x+y=4$. Đáp án D