Question

15．化简：a${\;}^{\frac{1}{3}}$+（a${\;}^{\frac{1}{3}}$-2b${\;}^{\frac{1}{2}}$）÷（a${\;}^{-\frac{2}{3}}$-$\frac{2\root{3}{b}}{a}$）×$\frac{\sqrt{a•\root{3}{{a}^{2}}}}{\root{5}{\sqrt{a}•\root{3}{a}}}$．

Answer 1

分析 利用分数指数幂的性质和运算法则直接求解即可．

解答 解：a${\;}^{\frac{1}{3}}$×（a${\;}^{\frac{1}{3}}$-2b${\;}^{\frac{1}{3}}$）÷（a${\;}^{-\frac{2}{3}}$-$\frac{2\root{3}{b}}{a}$）×$\frac{\sqrt{a•\root{3}{{a}^{2}}}}{\root{5}{\sqrt{a}•\root{3}{a}}}$
=${a}^{\frac{1}{3}}$×$\frac{{a}^{\frac{1}{3}}-2{b}^{\frac{1}{3}}}{{a}^{-\frac{2}{3}}-2{a}^{-1}{b}^{\frac{1}{3}}}$×$\frac{{a}^{\frac{1}{2}}•{a}^{\frac{1}{3}}}{{a}^{\frac{1}{10}}•{a}^{\frac{1}{15}}}$
=${a}^{\frac{1}{3}}$×$\frac{{a}^{\frac{4}{3}}-2a{b}^{\frac{1}{3}}}{{a}^{\frac{1}{3}}-2{b}^{\frac{1}{3}}}$×a${\;}^{\frac{2}{3}}$
=a²．

点评 本题考查有理数指数幂的化简求值，是基础题，解题时要认真审题，注意分数指数幂的性质和运算法则的合理运用．