Cho 

B

- Hướng dẫn giải

Ta có:

 $$\begin{gathered} {\log }_{2}x = 2{\log }_2\sqrt[3]{a^2} + 3{\log }_2\frac{1}{b^2\sqrt{b}}(a;b>0), x>0 \\ \iff {\log }_{2}x={\log }_2{a}^{\frac{4}{3}}+{\log }_2{b}^{-\frac{15}{2}} \\ \iff {\log }_{2}x={\log }_2{a}^{\frac{4}{3}}.b^{-\frac{15}{2}} \\ \iff {\log }_{2}x={\log }_2{a}^{\frac{4}{3}}.b^{-\frac{15}{2}} \\ \iff x = a^{\frac{4}{3}}.b^{-\frac{15}{2}} \end{gathered}$$

Do đó $x=a^{\frac{4}{3}}.b^{\frac{-15}{2}}$

Chọn B