Question

8．已知a，b∈R⁺，则$\frac{{\sqrt{{a^3}b}}}{{\root{3}{ab}}}$=（　　）

• A． ${a^{\frac{1}{6}}}{b^{\frac{7}{6}}}$
• B． ${a^{\frac{7}{6}}}{b^{\frac{1}{6}}}$
• C． ${a^{\frac{1}{3}}}{b^{\frac{1}{6}}}$
• D． ${a^{\frac{1}{2}}}{b^{\frac{1}{6}}}$

Answer 1

分析 利用根式与分数指数幂化简$\frac{{\sqrt{{a^3}b}}}{{\root{3}{ab}}}$=$\frac{{a}^{\frac{3}{2}}{b}^{\frac{1}{2}}}{{a}^{\frac{1}{3}}{b}^{\frac{1}{3}}}$=${a}^{\frac{3}{2}-\frac{1}{3}}$${b}^{\frac{1}{2}-\frac{1}{3}}$，从而解得．

解答 解：$\frac{{\sqrt{{a^3}b}}}{{\root{3}{ab}}}$=$\frac{{a}^{\frac{3}{2}}{b}^{\frac{1}{2}}}{{a}^{\frac{1}{3}}{b}^{\frac{1}{3}}}$
=${a}^{\frac{3}{2}-\frac{1}{3}}$${b}^{\frac{1}{2}-\frac{1}{3}}$
=${a^{\frac{7}{6}}}{b^{\frac{1}{6}}}$，
故选B．

点评 本题考查了根式与分数指数幂的互化．